All categories

3 years ago

A 15 ft. ladder leans against a wall. The

1 answer:

7 0

To find the angle between the the sides that equals 5 and 15, do the converse of sin.

Converse sin (5/15) ≈ 70.528
So the answer is C.

You might be interested in

Maddie loves music. ⅚ of her collection is hip-hop. ¼ of the remaining music is country. What fraction of her music collection i

Vlad [161]

If 5/6 of her collection is hip-hop, then 1/6 of her collection is not hip-hop. So 1/4 of the remaining music is country so 1/6 times 1/4 is 1/24.

3 0

3 years ago

Read 2 more answers

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

4 years ago

The number of fatalities due to car crashes, based on the number of miles driven, begins to climb after the driver is past age 6

andriy [413]

Answer:

The fataly rate per 100 million vehicle miles driven is approximately for average driver in the 50-54 age group is 0.7.

The fataly rate per 100 million vehicle miles driven is approximately for average driver in the 60-64 age group is 3.3268.

Step-by-step explanation:

The number of fatalities per 100 million vehicle miles driven is approximately given by the following function:

$N(x) = 0.3336x^{3} - 0.118x^{2} + 0.215x + 0.7$

In which x corresponds to the age group.

What is the fatality rate per 100 million vehicle miles driven for an average driver in the 50-54 age group?

This is N(0), since $x = 0$ represents those aged 50-54.

So:

$N(x) = 0.3336x^{3} - 0.118x^{2} + 0.215x + 0.7$

$N(0) = 0.7$

The fataly rate per 100 million vehicle miles driven is approximately for average driver in the 50-54 age group is 0.7.

In the 60-64 age group?

This is N(2), since $x = 2$ represents those aged 60-64.

$N(x) = 0.3336x^{3} - 0.118x^{2} + 0.215x + 0.7$

$N(2) = 0.3336*2^{3} - 0.118*2^{2} + 0.215(2) + 0.7 = 3.3268$

The fataly rate per 100 million vehicle miles driven is approximately for average driver in the 60-64 age group is 3.3268.

3 0

3 years ago

Which expression can be modeled using the number line

Anna [14]

Answer:

7 divided by 7/10 is modeled

5 0

3 years ago

Tom deposits $300 into an account that pays simple interest at a rate of 4% per year. How much interest will he be paid in the f

seropon [69]

Answer:

22.90 is the answer i can't write somet

use my computer has problems

I = P*r*t = 800*0.06*2 =

8 0

3 years ago