The weather report says the temperature is 20°C and will drop 5°C per hour for the next 6 hours. Daryl plans to begone for at least 6 hours, and he has a plant outside. If he wants the plant to remain in temperatures above -10°C,
should Daryl move his plant to a warmer location before leaving?
An inequality to model the problem is
The solution is​

Answer:

The radio station is at the center of a circle with an area of 40,000 square miles.  The furthest point that you could hear the broadcast would be on the outer edge of the circle or at the end of the radius of the circle.  The formula for the area of a circle is:

Area = π

So we take what we know and work the formula in reverse.

40,000 = πr²

40,000 / π = r²

12,732.4 = r²

√12,732.4 = √r²

112.84 = r

The radius of a circle with an area of 40,000 square miles is 112.84 miles.  This is the furthest point that you would be able to hear the broad cast.

Step-by-step explanation:

Answer:

 The correct answer is:  [B]:  " 0.56 " .

Step-by-step explanation:

Note:

 " 56%   = " " ;

              =  (56) ÷ (100)  =  0.56  ;  

  →  which corresponds to:  " Answer choice:  [B]: " 0.56 " .

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The amount of chlorine needed to treat a swimming pool is directly proportional to the volume of the poolWhat is the constant of proportionality for this relationship

(a) For n = 6, CL = 90%,  

The degrees of freedom: 5, Critical t-value: 2.571

(b) For n = 21, CL = 98%,

The degrees of freedom: 20, Critical t-value: 2.845

(c) For n = 29, CL = 95%,

The degrees of freedom: 28, Critical t-value: 2.048

(d) For n = 12, CL = 99%,

The degrees of freedom: 11, Critical t-value: 3.106

Use the concept of critical t- value defined as:

A critical value is a number that is used in hypothesis testing to compare to a test statistic and evaluate whether or not the null hypothesis should be rejected. The null hypothesis cannot be rejected if the test statistic's value is less extreme than the crucial value.

(a) Given that,

n = 6 and a confidence level of 90%,

The degrees of freedom are,

n-1 = 6-1

The degrees of freedom = 5.

To find the critical t-value,

Look it up in the t-distribution table using a confidence level of 90% and a degree of freedom of 5.

From the table,

The critical t-value is approximately 2.571.

(b) Given that,

n = 21 and a confidence level of 98%,

The degrees of freedom are,

n-1 = 21-1

The degrees of freedom = 20.

By referring to the t-distribution table with a confidence level of 98% and degrees of freedom of 20,

The critical t-value is approximately 2.845.

(c) Given that,

n = 29 and a confidence level of 95%,

The degrees of freedom are,

n-1 = 29-1

The degrees of freedom = 28

Using the t-distribution table with a confidence level of 95% and degrees of freedom of 28,

The critical t-value is approximately 2.048.

(d) Given that,

n = 12 and a confidence level of 99%,

The degrees of freedom are,

n-1 = 12-1

The degrees of freedom = 11

By consulting the t-distribution table with a confidence level of 99% and degrees of freedom of 11,

The critical t-value is approximately 3.106.

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Final answer:

To find the degrees of freedom and critical t-value for each given sample size and confidence level, we can use the t-distribution and a t-table. The degrees of freedom (df) for each sample is equal to the sample size minus 1. The critical t-value can be found using the t-table with the corresponding degrees of freedom and the confidence level.

Explanation:

To find the degrees of freedom and critical t-value for each given sample size and confidence level, we can use the t-distribution and a t-table. The degrees of freedom (df) for each sample is equal to the sample size minus 1. For example, for (a) n = 6, df = 6 - 1 = 5. The critical t-value can be found using the t-table with the corresponding degrees of freedom and the confidence level.

For (a) n = 6, CL = 90%, the critical t-value is approximately 1.943.

For (b) n = 21, CL = 98%, the critical t-value is approximately 2.861.

For (c) n = 29, CL = 95%, the critical t-value is approximately 2.045.

For (d) n = 12, CL = 99%, the critical t-value is approximately 3.106.

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