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3 years ago

14

A square has an area of approximately 500 square feet. Which of the following would be the most reasonable estimate of the lengt

h of the side of the square?

1 answer:

vlabodo [156]3 years ago

8 0

Answer:

22.36

Step-by-step explanation:

22.36 X 22.36 =(Approx) 500 feet2

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Solve AABC. Round your answers to the nearest hundredth, if necessary

Aloiza [94]

Answer:

$C=25^{\circ},\\a\approx 10.72,\\b\approx 11.83$

Step-by-step explanation:

The sum of the interior angles of a triangle is 180 degrees. Thus, angle C must be $180-90-65=25^{\circ}$ .

In any triangle, the Law of Sines is given by $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$ .

Therefore, we have:

$\frac{\sin 90^{\circ}}{b}=\frac{\sin 25^{\circ}}{5},\\\\b=\frac{5\sin90^{\circ}}{\sin 25^{\circ}}=11.8310079158\approx \boxed{11.83}$

$\frac{a}{\sin 65^{\circ}}=\frac{5}{\sin25^{\circ}},\\a=\frac{5\sin 65^{\circ}}{\sin25^{\circ}}=10.7225346025\approx \boxed{10.72}$

8 0

3 years ago

Swimsuit discounted 65% on sale for $20.30.

Dafna1 [17]

Original price of the swimsuit = x

If something was discounted 65%, that means you are paying 35% for it.
(100% - 65% = 35% or 0.35)

x(0.35) = 20.30

divide both sides by 0.35

x = 58

The swimsuit was originally 58 dollars.

8 0

3 years ago

Read 2 more answers

The navy reports that the distribution of waist sizes among male sailors is approximately normal, with a mean of 32.6 inches and

grandymaker [24]

Answer:

a) 87.49%

b) 2.72%

Step-by-step explanation:

Mean of the waist sizes = u = 32.6 inches

Standard Deviation of the waist sizes = $\sigma$ = 1.3 inches

It is given that the Distribution is approximately Normal, so we can use the z-distribution to answer the given questions.

Part A) A male sailor whose waist is 34.1 inches is at what percentile

In order to find the percentile score of 34.1 we need to convert it into equivalent z-scores, and then find what percent of the value lie below that point.

So, here x = 34.1 inches

The formula for z score is:

$z=\frac{x-u}{\sigma}$

Using the values in the above formula, we get:

$z=\frac{34.1-32.6}{1.3}=1.15$

Thus, 34.1 is equivalent to z score of 1.15

So,

P( X ≤ 34.1 ) = P( z ≤ 1.15 )

From the z-table we can find the probability of a z score being less than 1.15 to be: 0.8749

Thus, the 87.49 % of the values in a Normal Distribution are below the z score of 1.15. For our given scenario, we an write: 87.49% of the values lie below 34.1 inches.

Hence, the percentile rank of 34.1 inches is 87.49%

Part B)

The regular measure of waist sizes is from 30 to 36 inches. Any measure outside this range will need a customized order. We need to find that what percent of the male sailors will need a customized pant. This question can also be answered by using the z-distribution.

In a normal distribution, the overall percentage of the event is 100%. So if we find what percentage of values lie between 30 and 36, we can subtract that from 100% to obtain the percentage of values that are outside this range and hence will need customized pants.

First step is again to convert the values to z-scores.

30 converted to z scores will be:

$z=\frac{30-32.6}{1.3}=-2$

36 converted to z score will be:

$z=\frac{36-32.6}{1.3}=2.62$

So,

P ( 30 ≤ X ≤ 36 ) = P ( -2 ≤ z ≤ 2.62 )

From the z table, we can find P ( -2 ≤ z ≤ 2.62 )

P ( -2 ≤ z ≤ 2.62 ) = P(z ≤ 2.62) - P(z ≤ -2)

P ( -2 ≤ z ≤ 2.62 ) = 0.9956 - 0.0228

P ( -2 ≤ z ≤ 2.62 ) = 0.9728

Thus, 97.28% of the values lie between the waist sizes of 30 and 36 inches. The percentage of the values outside this range will be:

100 - 97.28 = 2.72%

Thus, 2.72% of the male sailors will need custom uniform pants.

The given scenario is represented in the image below. The black portion under the curve represents the percentage of male sailors that will require custom uniform pants.

3 0

4 years ago

Pic is please please else\

Zanzabum

I can't see the picture clearly sorry I wish I can help…

3 0

3 years ago

Select the correct answer

xxMikexx [17]

Answer:

Option B. 220

Step-by-step explanation:

Let

x ----> the number of student tickets sold

y ----> the number of adult tickets sold

we know that

$x+y=360$

$y=360-x$ ----> equation A

$3x+5y=1,360$ ----> equation B

Solve the system by substitution

substitute equation A in equation B

$3x+5(360-x)=1,360$

solve for x

$3x+1,800-5x=1,360$

$5x-3x=1,800-1,360$

$2x=440$

$x=220\ student\ tickets$

5 0

3 years ago

Read 2 more answers