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3 years ago

The perimeter of a rectangle is 56 feet and

Mathematics

1 answer:

Marat540 [252]3 years ago

4 0

Answer:

Step-by-step explanation:

P = 2(L + W)

Area = L*W

Area = 192

(L + W)*2 = 56

L+W = 28

L = 28 - W

W*(28 - W) = 192

28W - w^2 = 92

-w^2 + 28w - 192 = 0

w^2 - 28w + 192 = 0

This factors into

(w - 12)(w - 16) = 0

w - 12 = 0

w = 12

L = 28 - 12 = 16

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In ΔQRS, the measure of ∠S=90°, the measure of ∠Q=31°, and RS = 27 feet. Find the length of SQ to the nearest tenth of a foot.

Soloha48 [4]

Answer:

44.9

Step-by-step explanation:

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3 years ago

Martha has $20 to spend on a t-shirt. She has a coupon for 20% off. The t-shirt regularly sells for $22.50 plus 10% sales tax.

rodikova [14]

Answer:

1 and 3

Step-by-step explanation:

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3 years ago

11.50 + 1.15x = 12.65

Illusion [34]

Answer: the answer is x= 1

Step-by-step explanation:

So we are dealing with 11.50 - 11.50 + 1.15x = 12.65 - 11.50

our first step is Simplify both sides of the equation so:

11.5 + -11.50 + 1.15x = 12.65 + -11.50 turns into:

(1.15x) + (11.5+ -11.5)= (12.65 + -11.5)

**we know 11.5 + -11.5 will cancel each other since they are opposite**

so it will be 1.15x= 1.15

now the very last step is to divide 1.15x on both sides which gives us 1

7 0

3 years ago

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What do i do? What do they mean ...HELP ME!!

katrin2010 [14]

Sin x = a/c

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4 0

3 years ago

Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

blsea [12.9K]

Answer:

a) Percentage of students scored below 300 is 1.79%.

b) Score puts someone in the 90th percentile is 638.

Step-by-step explanation:

Given : Suppose a student's score on a standardize test to be a continuous random variable whose distribution follows the Normal curve.

(a) If the average test score is 510 with a standard deviation of 100 points.

To find : What percentage of students scored below 300 ?

Solution :

Mean $\mu=510$ ,

Standard deviation $\sigma=100$

Sample mean $x=300$

Percentage of students scored below 300 is given by,

$P(Z\leq \frac{x-\mu}{\sigma})\times 100$

$=P(Z\leq \frac{300-510}{100})\times 100$

$=P(Z\leq \frac{-210}{100})\times 100$

$=P(Z\leq-2.1)\times 100$

$=0.0179\times 100$

$=1.79\%$

Percentage of students scored below 300 is 1.79%.

(b) What score puts someone in the 90th percentile?

90th percentile is such that,

$P(x\leq t)=0.90$

Now, $P(\frac{x-\mu}{\sigma} < \frac{t-\mu}{\sigma})=0.90$

$P(Z< \frac{t-\mu}{\sigma})=0.90$

$\frac{t-\mu}{\sigma}=1.28$

$\frac{t-510}{100}=1.28$

$t-510=128$

$t=128+510$

$t=638$

Score puts someone in the 90th percentile is 638.

5 0

3 years ago