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3 years ago

Need help please help

Mathematics

2 answers:

professor190 [17]3 years ago

5 0

153.9 inches is the correct answer

Ann [662]3 years ago

3 0

Answer:

153.9 in²

Step-by-step explanation:

Using the formula: A = π r²

Where r = the radius and π is 3.14.

The radius is the diameter divided by 2.

14 divided by 2 is 7.

Plug in 7 into the formula:

A = π 7²

7² is 49.

49 times π is equal to 153.86.

Rounding this up to the nearest tenth, you'll get 153.9

You might be interested in

Decide whether the two expressions are equivalent for each problem below.

sashaice [31]

C and d are equivalent

3 0

3 years ago

Manuel se ha gastado 2/8 partes de la paga en ropa y 1/12 parte en dulces. ¿Qué fracción de la paga le queda?, ¿Si le dieron 45€

Sindrei [870]

Responder:

2/3; 30 €

Explicación paso a paso:

Fracción gastada en ropa = 2/8

Fracción gastada en dulces = 1/12

Fracción restante:

1 - (Fracción gastada en ropa + fracción gastada en dulces)

1 - (2/8 + 1/12)

L. C. M de 8 y 12 = 24

1 - (6 + 2) / 24

1 - 8/24

1 - 1/3

= (3 - 1) / 3

= 2/3

Si la paga era de 45;

Cantidad restante: 2/3 de 45

= 90/3

= 30 €

8 0

2 years ago

A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se

bearhunter [10]

Answer:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

Step-by-step explanation:

We know the following info from the problem

$\bar X_1 = 14.5$ sample mean for the group 1

$s_1 = 3.98$ the standard deviation for the group 1

$n_1= 20$ the sample size for group 1

$\bar X_2 = 13.9$ sample mean for the group 2

$s_1 = 4.03$ the standard deviation for the group 2

$n_2= 17$ the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given :

$df = n_1 +n_2- 2 = 20+17-2=35$

The confidence level is 0.9 or 90% and the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 = 0.05$ and the critical value for this case is:

$t_{\alpha/2} = 1.69$

And replacing the info given we got:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

5 0

2 years ago

Which fraction is NOT equivalent to 4/9 ?

BARSIC [14]

Everything that is not 4/9

8 0

3 years ago

Read 2 more answers

What is -3 - 9n= 42 I can’t be negative

maw [93]

If you were to correctly simplify this equation, the answer would be a negative.

The answer is: n = -5

Here's how I got my answer:

Step 1: Add 3 to both sides. Which leaves us with $-9n=42+3$

Step 2: Simplify $42+3$ to 45. Which leaves us with $-9n = 45$

Step 3: Divide both sides by -9. Which leaves us with $n = -45/9$

Step 4: Simplify $-45/9$ , to get your answer, n = -5.

4 0

3 years ago