The base of a triangle is four times its height.If the area of the triangle is 242 square millimeters,find its height

A case of water costs $3.80. If Safeway marks up the price 10%, what is the new price of the

anyanavicka [17]

Answer:

$4.18

Step-by-step explanation:

You take $3.80 and multiply by it .1 (turn 10% into a decimal by moving the decimal to the right two places).

This gives you .38

Then you add .38 to $3.80 which gives you $4.18.

5 0

3 years ago

A piece of fabric is 1/3 yard by 3 5/8 yards. Irma solves 1/3⋅3 5/8 to find the area of the fabric in square yards. Enter number

Harlamova29_29 [7]

For this case we have:

The area of the piece of fabric is given by $A = a * b$

Where

$a = \frac{1}{3}yards$

$b = 3 \frac{5}{8}yards$

Substituting in the formula of the area we have:

$A = \frac{1}{3}* 3 \frac{5}{8}\\$

$A = \frac{1}{3}* (3 + \frac{5}{8})\\$

$A = 1 + \frac{5}{24}$

$A = 1\frac{5}{24}yards ^ 2\\$

Thus, the area of the piece of fabric is given by: $A = 1\frac{5}{24}yards ^ 2\\$

Answer:

$A = 1\frac{5}{24}yards ^ 2\\$

4 0

3 years ago

The population of Tinyville is 16.2 Tinys per acre. Exactly 405 Tinys live in Tinyville. How many acres is Tinyville? Only enter

Mice21 [21]

The answer is 25 because 405/16.2=25.

8 0

3 years ago

Write the algebraic expression described. Recall that two angles are supplementary if their sum is 180 degrees. If one angle mea

Alina [70]

If one angle is d° then it's supplement is
(180-d)°

8 0

3 years ago

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work i

USPshnik [31]

Answer:

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

$n = 55, \pi = \frac{24}{55} = 0.4364$

93% confidence level

So $\alpha = 0.07$ , z is the value of Z that has a pvalue of $1 - \frac{0.07}{2} = 0.965$ , so $Z = 1.81$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 - 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.3154$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.4364 + 1.81\sqrt{\frac{0.4364*0.5636}{55}} = 0.5574$

The 93% confidence interval for the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574). This means that we are 93% sure that the true proportion of masks of this type whose lenses would pop out at 325 degrees is (0.3154, 0.5574).

8 0

3 years ago