A wildlife biologist catches and releases 20 fish from two different lakes at random locations. he catches 10 fish at lake palmer and 10 fish at lake dalton. he measures the length of each fish to the nearest quarter of an inch. palmer dalton first quartile 6.75 5.5 second quartile (median) 10.25 6.75 third quartile 13 7.5 based on the samples, what generalization can be made? at least 25 percent of the fish in both lakes are no longer than 6 inches. the interquartile range for lake dalton is 2 inches greater than the interquartile range for lake palmer. the first quartile value at lake palmer is 1.25 inches longer than the first quartile value at lake dalton. not enough information is provided to draw any of these conclusions.

The greatest common factor of [tex]\(15x^2y^3\)[/tex] and [tex]\(-20x^3yz\)[/tex] is [tex]\(5x^2y\)[/tex].

To find the greatest common factor (GCF) of [tex]\(15x^2y^3\)[/tex] and [tex]\(-20x^3yz\)[/tex], we need to identify the highest power of each common factor that both terms share.

List the prime factors of each term:

[tex]\(15x^2y^3 = 3 \times 5 \times x^2 \times y^3\)[/tex]

[tex]\(-20x^3yz = -1 \times 2 \times 2 \times 5 \times x^3 \times y \times z\)[/tex]

Identify the common factors:

Common factors: [tex]\(5\), \(x^2\), \(y\)[/tex]

Find the lowest exponent for each common factor:

For (5), it appears in both terms, so we use the smaller exponent, which is [tex]\(5^1\)[/tex].

For [tex]\(x^2\)[/tex], it appears in both terms, so we use the smaller exponent, which is [tex]\(x^2\)[/tex].

For (y), it appears in both terms, so we use the smaller exponent, which is [tex]\(y^1\)[/tex].

Multiply the common factors:

[tex]\(GCF = 5 \times x^2 \times y = 5x^2y\).[/tex]

So, the greatest common factor of [tex]\(15x^2y^3\)[/tex] and [tex]\(-20x^3yz\)[/tex] is [tex]\(5x^2y\)[/tex].

4:5 ≡ 8: 10 is the correct way to compares these ratios.

" Equivalent means when two or more things have same value but different expression."

According to the question,

Ratio of girls to boys in Mr. Hansen's class = 4 : 5

Ratio of girls to boys in Ms. Luna's class = 8 : 10

Compare both the ratio's we get,

As

[tex]\frac{8}{10} =\frac{2}{2} * \frac{4}{5}[/tex]

Therefore,

4 : 5 ≡ 8 : 10

Hence, 4:5 ≡ 8: 10 is the correct way to compares these ratios.

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Answer:

67,600

Step-by-step explanation:

10 gallons of 30% juice was mixed with 90 gallons of 10% juice to make 100 gallons of 12% juice.

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the amount of 30% juice and y represent the amount of 10% juice, hence:

x + y = 100         (1)

Also:

0.3x + 0.1y = 0.12(100)         (2)

From eqn 1 and 2:

x = 10, y = 90

10 gallons of 30% juice was mixed with 90 gallons of 10% juice to make 100 gallons of 12% juice.

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To find when the building will be completely depreciated, you solve for x in the equation y = 360,000 - 1,500x by setting y to 0. The solution is x = 240, meaning the building will be totally depreciated after 240 months or 20 years.

In this case, you are looking to find the value of x (the number of months) when the value of the building (y) becomes 0. The equation given is y = 360,000 - 1,500x. Set y = 0 and solve for x

So, the building will be completely depreciated after 240 months or 20 years.

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Let us say that m is the number of months. So that the equations for option 1 (O1) and option 2 (O2) are:

O1 = 30 m + 100

O2  = 25 m + 150

TO determine this, we equate O1 and O2 so that:

30 m + 100 = 25 m + 150

5 m = 50

m = 10

So at the 10th month, the cost for both options are equal. So we select month 11.

Answer:

D) month 11

Ria will need 2677920 Rupees to pay her tuition fees

Data;

We need to convert the currency from dollars to rupees

From the data given in the question, $1 is equivalent to 63.7 Rupees.

This information can further be used to determine how many Rupees she needs to pay her Tuition fees.

[tex]\$1 = 63.76 Rupees\\\$42,000 = x\\x = 42000 * 63.76\\x = 2677920[/tex]

Ria will need 2677920 Rupees to pay her tuition fees

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The 95% confidence interval for the mean wrist circumference for all men is 18.91 to 19.69 centimeters.

To calculate the confidence interval for the mean wrist circumference, we can use the formula: CI = x ± t(s/√n), where CI is the confidence interval, x is the sample mean, t is the t-value corresponding to the desired confidence level, s is the sample standard deviation, and n is the sample size.

In this case, the sample mean is 19.3 centimeters, the sample standard deviation is 1.2 centimeters, and the sample size is 48. We can find the t-value for a 95% confidence level using a t-table or calculator. Assuming a normal distribution, the t-value is approximately 1.96. Plugging in these values into the formula, we get:

CI = 19.3 ± 1.96(1.2/√48) = 19.3 ± 0.3956

Therefore, the 95% confidence interval for the mean wrist circumference is approximately 18.91 to 19.69 centimeters.

Answer:

[tex]x=-\dfrac{11}{6},\dfrac{8}{7}[/tex]

Step-by-step explanation:

Given: [tex](7x-8)(6x+11)[/tex]

Set the expression to 0 to solve for x

[tex](7x-8)(6x+11)=0[/tex]

Set each factor to 0 and solve for x

7x-8 = 0    or    6x+11 = 0

7x = 8      or    6x = -11

[tex]x=\dfrac{8}{7}[/tex]    or   [tex]x=-\dfrac{11}{6}[/tex]

hence, The values of x are [tex]x=-\dfrac{11}{6},\dfrac{8}{7}[/tex]

Answer: 21 quarts

Step-by-step explanation: In this problem, we are asked to convert 5.25 gallons into quarts. To convert gallons into quarts, we use the conversion factor for gallons and quarts.

Conversion factor ⇒ 4 quarts = 1 gallon

Next, notice that we are going from a larger unit "gallons" to a smaller unit "quarts" and when we go from a larger unit to a smaller unit we multiply.

We will multiply 5.25 by the conversion factor which is 4 to get 21 quarts.

Therefore, 5.25 gallons is equal to 21 quarts.

This problem is simply a multiplication problem. All we have to do in this case is to multiply 0.24 by 13, or the other way around multiply 13 by 0.24. We can do this in the calculator which gives us:

13 x 0.24 = 3.12

Answer:

A. 3.12

Final answer:

The proportion of a normal distribution for z scores greater than 1.04 is 0.1492, which is found by subtracting the area to the left (0.8508) from the total area under the curve (1).

Explanation:

To find the proportion of a normal distribution that corresponds to z scores greater than 1.04, we must look up the area to the left of the z-score using a z-table or a standard normal distribution table. Once we have the area to the left, we can calculate the area to the right by subtracting this value from 1 (since the total area under the curve is 1). For a z-score of 1.04, the z-table shows that the area to the left is approximately 0.8508. Therefore, the proportion that is greater than a z-score of 1.04 is 1 - 0.8508, which equals 0.1492.

The standard normal distribution (Z~N(0,1)) has a total area of 1 under the curve, and it is symmetric about the mean, which is 0. Z-scores give us a way to determine how many standard deviations an observation is from the mean. Additionally, common z-scores used to determine confidence intervals or the proportion of observations within a certain range are tied to the percentage areas under the curve. For example, a z-score of ±1 encompasses approximately 68% of the data, ±2 about 95%, and ±3 approximately 99.7%.