Answer: False
Martin is asked to find the probability of getting one head and three tails on for coin tosses this is a simple event.
A. True
B. False
Find the y-intercept and x-intercept of the line.
5x - 4y = 15
y-intercept: __
x-intercept: __
∠EFG and ∠GFH are a linear pair, m∠EFG=3n+19, and m∠GFH=55+33 What are m∠EFG and m∠GFH?
How many sixteenth notes would be needed to have the same duration as 3 quarter notes? Represent this as a fraction
(6.3 × 1011) ÷ (7 × 105)
can someone help me please
Explain how you would use a number line to find the absolute value of –12.
which statement is true about angles 1 and 2
Answer:
Option a) They are adjacent angles
Step-by-step explanation:
Given is a picture of a graph with angles 1 to 6 around it.
Out of these angle 2 and 5 are right angles.
1 and 2 are adjacent to each other.
They are not complementary because 1+2 not equals 90
They are neither supplementary since sum does not equal 180
They cannot be vertical because they are not formed by intersection of two lines.
Hence only option a is right
Option a) They are adjacent angles
how much would $500 invested at 6% interest compounded annually be worth after 4 years
A rectangle has a base of 3 inches and a height of 9 inches. If the dimensions are doubled, what will happen to the area of the rectangle?
Answer:
Area will increase by 4 times
Step-by-step explanation:
Given: A rectangle has a base of 3 inches and a height of 9 inches.
To find: If the dimensions are doubled, what will happen to the area of the rectangle?
Solution:
It is given that a rectangle has a base of 3 inches and a height of 9 inches.
Now, to find if the dimensions are doubled what will happen to the area, first we need to find the original area.
Original area of rectangle, when base is 3 inches and height 9 inches is
[tex]9\times3=27[/tex] square inches
Now, when dimensions are doubled , the base becomes 6 inches and height becomes 18 inches
So, new area becomes [tex]6\times18=108[/tex] square inches.
Now,
[tex]\frac{\text{new area}}{\text{original area} }=\frac{108}{27}[/tex]
[tex]\implies\frac{\text{new area}}{\text{original area} }=\frac{4}{1}[/tex]
Hence, the area will increase by 4 times.
Five students visiting the student health center for a free dental examination during national dental hygiene month were asked how many months had passed since their last visit to a dentist. their responses were as follows. 5 18 12 24 28 assuming that these five students can be considered a random sample of all students participating in the free checkup program, construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program. (give the answer to two decimal places.)
To find for the value of the confidence interval, let us first calculate for the values of x and s, the mean and standard deviation respectively.
x = (5 + 18 + 12 + 24 + 28) / 5
x = 17.4 months
s = sqrt{[(5 – 17.4)^2 + (18 – 17.4)^2 + (12 – 17.4)^2 + (24 – 17.4)^2 + (28 – 17.4)^2]/(5-1)}
s = 9.21
The formula for the confidence interval is given as:
Confidence Interval = x ± t s / sqrt(n)
Where t can be taken from standard distribution tables at 95% level at degrees of freedom = n – 1 = 4, t = 2.132. Therefore:
Confidence Interval = 17.4 ± 2.132 * 9.21 / sqrt(5)
Confidence Interval = 17.4 ± 8.78
Confidence Interval = 8.62 months, 26.18 months
Final answer:
To construct a 95% confidence interval for the mean number of months elapsed since the last visit to a dentist, use the sample mean and sample standard deviation. Calculate the confidence interval using the formula and given data.
Explanation:
In order to construct a confidence interval for the mean number of months elapsed since the last visit to a dentist for the population of students participating in the program, we can use the sample mean and sample standard deviation. The formula for constructing a confidence interval for the mean is:
Confidence Interval = sample mean ± (critical value) × (sample standard deviation / √sample size)
Using the given data, the sample mean is 17.4 months and the sample standard deviation is 9.18 months. With a confidence level of 95%, the critical value is 2.776. Plugging these values into the formula, we get:
Confidence Interval = 17.4 ± (2.776) × (9.18 / √5)
Calculating this, we find that the 95% confidence interval for the mean number of months elapsed since the last visit to a dentist is approximately 4.81 to 29.99 months.
What is the correct name for this circle?
a bell tolls every 10 minutes. another bell tolls every 15 minutes. both bells toll at 6:00pm. they will toll together at what time
Both the bells will ring after 30 minutes at 6:30 pm.
What is the lowest common multiple?The smallest common positive number that is a multiple of two or more numbers.
Given that, a bell tolls every 10 minutes. another bell tolls every 15 minutes. both bells toll at 6:00pm.
To find the time for which both will ring together, we will find LCM of 10and 15
10 = 5x2
15 = 5x3
LCM = 5x2x3 = 30
The LCM of 15 and 10 is 30
So, both will ring after 30 minutes.
Right now the time is 6:00 pm, so after 30 minutes it will be 6:30 pm
Hence, Both the bells will ring after 30 minutes at 6:30 pm.
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Find all values of $x$ such that $6= \dfrac{35}{x} -\dfrac{49}{x^2}$. If you find more than one value, then list your solutions in increasing order, separated by commas.
Answer:
[tex]x=\frac{7}{3} , \frac{7}{2}[/tex]
Step-by-step explanation:
[tex]6= \frac{35}{x} - \frac{49}{x^2}[/tex]
Now we need to solve for x
To get 'x' alone we make the denominators same
LCD = x^2
WE multiply the whole equation by x^2
[tex]6x^2 = 35x - 49[/tex]
Now we the equation =0, move all the terms to left hand side
[tex]6x^2-35x + 49=0[/tex]
Now we apply quadratic formula to solve for x
a= 6, b= -35 , c= 49
[tex]x= \frac{-b+-\sqrt{b^2-4ac}}{2a}[/tex]
[tex]x= \frac{-(-35)+-\sqrt{(-35)^2-4(6)(49)}}{2*6}[/tex]
[tex]x= \frac{35+-\sqrt{49}}{12}[/tex]
[tex]x= \frac{35+-7}{12}[/tex]
Now frame two equations , one with + and another with -
[tex]x= \frac{35+7}{12}[/tex] [tex]x= \frac{35-7}{12}[/tex]
[tex]x= \frac{42}{12}[/tex] [tex]x= \frac{28}{12}[/tex]
[tex]x= \frac{7}{2}[/tex] [tex]x= \frac{7}{3}[/tex]
So value of x= {7/3, 7/2}
Find the product of (x − 2i)^2.
The product of (x-2i)²=x²-4+i4x
What is the process of multiplication of complex numbers?The product or multiplication of two complex numbers is also a complex number. The formula for multiplying complex numbers is: (a + ib) (c + id) = (ac - bd) + i(ad + bc).
Given here (x-2i)² =(x-2i) (x-2i)
=x²-4+i4x
Hence, the product is x²-4+i4x
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bacterial colonies can triple in size every 4 days. if you start with 150 bacteria microorganisms, how large would the colony be after 16 days?
type the slope-intercept equation of the line that passes through the points (0,2) and (2,0) y={?]x+{ }
A cold front moved in last weekend. In 8 hours overnight, the temperature outside dropped from 14 degrees to -10. What was the average temperature change for each hour?
I need the answer as quick as I can and I put it at max points!
Answer:
Step-by-step explanation:
The difference between 14 and 0 is 14, and the difference between 0 and -10 is 10. 14+10=24 for total change. For the average over 8 hours, we have 24/8=3 degrees
On monday eliza read her book. on tuesday she read three times as long as she read on monday. on wednesday she read 20 minutes less than tuesday. on thursday she read for 20 minutes which was half as long as she read on wednesday. how many minutes did eliza read over the 4-day period
Which point satisfies both f(x)=2^x and g(x)=3^x (0,1) (0,-1) (1,0) (-1,0)
Given: The coordinates of iscosceles trapezoid JKLM are J(-b, c), K(b,c), L(a,0), and M(-a,0).
Prove: The diagonals of an isosceles trapezoid are congruent.
As part of the proof, find the length of KM
A) a2+b2+c2
B) (-a+b)2+c2
C) (a+b)2+c2
Answer with explanation:
It is given that, coordinates of Isosceles trapezoid J K L M are J(-b, c), K(b,c), L(a,0), and M(-a,0).
To Prove: The diagonals of an isosceles trapezoid are congruent.
Proof:
Distance formula , that is distance between two points in x y plane is given by
[tex]=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
Where, [tex](x_{1},y_{1}),(x_{2},y_{2})}[/tex] are coordinates of two points in the plane.
Length of Diagonal J L
[tex]=\sqrt{(a+b)^2+(0-c)^2}\\\\=\sqrt{(a+b)^2+c^2}[/tex]
Length of Diagonal K M
[tex]=\sqrt{(a+b)^2+(0-c)^2}\\\\=\sqrt{(a+b)^2+c^2}[/tex]
So, we can see that,
J L = KM [tex]=\sqrt{(a+b)^2+(c)^2}[/tex]
Hence,The diagonals of an isosceles trapezoid are congruent.
So ,
[tex]KM=\sqrt{(a+b)^2+c^2}[/tex]
Option C
In triangle ABC, angle B = 30°, a = 210, and b = 164. The measure of angle A to the nearest degree is a0
The length of the third side of the triangle would be 106.5 units approx.
What is a triangle?A triangle is a two - dimensional figure with three sides and three angles.
The sum of the angles of the triangle is equal to 180 degrees.
∠A + ∠B + ∠C = 180°
Given is that in triangle ABC, angle B = 30°, a = 210, and b = 164.
The cosine formula is given as follows -
c² = a² + b² + 2abcos(α)
c = √(a² + b² + 2abcos(α))
c = √(210)² + (164)² + 2(210)(164)cos(30°)
c = 106.5 units approx.
Therefore, the length of the third side of the triangle would be 106.5 units approx.
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Mr. calloway is an algebra teacher. every class period he draws a piece of paper out of a hat without looking to determine the number of homework problems he will assign. each different color of paper represents a different number of homework problems. the hat contains 2 blue, 6 red, 10 yellow, and 7 purple pieces of paper.what is the probability that mr. calloway draws a purple piece of paper during the first class period and a blue piece of paper during the second class period if he replaces all pieces of paper before each drawing?
The probability that Mr. Calloway will draw a purple piece of paper during the first class period and a blue piece during the second class period with replacement is [tex]\frac{14}{25}[/tex].
The question involves calculating the probability of drawing a purple piece of paper during the first class period and a blue piece of paper during the second class period with replacement. To find this probability, we first calculate the probability of each event separately since the events are independent. Mr. Calloway's hat contains a total of 25 pieces of paper (2 blue, 6 red, 10 yellow, and 7 purple).
The probability of drawing a purple piece of paper in the first class period is the number of purple papers divided by the total number of papers:
P(purple) = Number of purple pieces / Total pieces = [tex]\frac{7}{25}[/tex]
Since the pieces are replaced, the probabilities in the second class period are the same as in the first. Thus, the probability of drawing a blue piece of paper in the second class is:
P(blue) = Number of blue pieces / Total pieces = [tex]\frac{2}{25}[/tex]
Because these events are independent, we multiply the probabilities of each event happening:
Total probability = P(purple) times P(blue) = ([tex]\frac{7}{25}[/tex]) times ([tex]\frac{2}{25}[/tex]) = [tex]\frac{14}{25}[/tex].
Therefore, the probability that Mr. Calloway draws a purple piece of paper during the first class period and a blue piece of paper during the second class period with replacement is [tex]\frac{14}{25}[/tex].
Use ABC to find the value of sin A. See picture below. Thanks!
Which Graph correctly represents x+2y≤4?
Answer with Step-by-step explanation:
We are given an inequality:
x+2y≤4
We have to determine its correct graph
In graph A and graph C (0,4) lies in shaded region but (0,4) does not satisfy the inequality(since, 0+2×4=8 which is not less than or equal to 4)
Hence, A and C are not the graph of this inequality
In graph D (0,2) does not lie in shaded area but it satisfies the inequalityHence, D is also not the graph of this inequality
Hence, correct graph of x+2y≤4 is:
Graph B
Perform the operation(s) and write the answer in simplest form.
( 1/2 + 1/3) / 15/4
A.) 3 1/8
B.) 53/90
C.) 2/9
D.) 8/75
Math help needed. Thank you
Julio just bought a $267,900 house. He had a 20 year mortgage with a fixed rate of 5.875%. Julio's monthly payments are $1558.09. What percent of the purchase price was Julio's down payment?
Answer: c
Step-by-step explanation:
16x^2=100
How many solutions will there be to the following equation?
Answer:
There are two solutions [tex]x=\frac{5}{2},-\frac{5}{2}[/tex]
Step-by-step explanation:
Given : Equation [tex]16x^2=100[/tex]
To find : How many solutions will there be to the following equation?
Solution :
Equation [tex]16x^2=100[/tex]
Solve the equation,
Divide by 16 both side,
[tex]\frac{16x^2}{16}=\frac{100}{16}[/tex]
[tex]x^2=\frac{100}{16}[/tex]
Taking root both side,
[tex]x=\sqrt{\frac{100}{16}}[/tex]
[tex]x=\sqrt{\frac{10^2}{4^2}}[/tex]
[tex]x=\pm\frac{10}{4}}[/tex]
[tex]x=\pm\frac{5}{2}}[/tex]
Therefore, There are two solutions [tex]x=\frac{5}{2},-\frac{5}{2}[/tex]
The Pyramid of Giza is one of the largest pyramid structures still standing in Egypt. It is a right pyramid with a square base, a base length of 230 m, and height of 150 m. The area of the base is __________ The volume is ________
Answer:- The area of base is 52900 square meters. The volume is 2654000 cubic meters.
Explanation:-
Base length of right pyramid (square) a =230m
Height of right pyramid = 150m
Area of base of right pyramid[tex]=a^2=(230)^2=52900\ m^2[/tex]
Volume of right pyramid with square base[tex]=\frac{1}{3}a^2h\\=\frac{1}{3}\times52900\times150=2645000\ m^3[/tex]
Thus, the area of base is 52900 square meters and the volume is 2654000 cubic meters.
The area of the base of the pyramid is 52900 m²
The volume of the square base pyramid of Giza is 2645000 m³
The pyramid is a square base pyramid. Therefore,
volume of a square base pyramid;v = 1 / 3 Bhwhere
B = base area
h = height
h = 150 m
Therefore,
area of the base = l²
where
l = length of side
area of the base = 230² = 52900 m²
Volume = 1 / 3 × 52900 × 150
volume = 7935000 / 3
volume = 2645000 m³
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Which of the following equations have graphs that are parallel to the graph of the equation y=-3/2x+8?
I. 3x + 2y = 10
II. 2x − 3y = 9
III. 6x + 4y = 28
IV. 3x − 2y = 8
I and III only
II and III only
IV only
III only