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3 years ago

I need help with this problem! I'll give brainlist!

Mathematics

1 answer:

aliya0001 [1]3 years ago

4 0

Step-by-step explanation:

use CAH method (Cos angle = adjacent/hypotenuse)

Given adjacent = 5

Hypotenuse = 12,

$\cos(x) = \frac{adjacent}{hypotenuse} \\ \cos(x) = \frac{5}{12} \\ x = {cos}^{ - 1} ( \frac{5}{12} ) \\ = 65.4deg(nearest \: tenth)$

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Since 20% or 1/5, is ___, there will be ___ word problems on the test

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Step-by-step explanation:

One fifth of fifty is 10.

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3 years ago

How does place value help me divide?

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4 years ago

The data below represent the ages of people working at a store.

Slav-nsk [51]

Answer:

The first step is sort the values on increasing order and we got:

18, 18, 19, 19, 19, 20, 22, 23. 25, 26, 31

And in order to find the first quartile we can separate the first 5 lowest values from the largest 5 values. And for this case we focus on these values:

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And if we find the median of the last 5 values we can find the first quartile and we got:

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The best answer would be:

Step-by-step explanation:

For this problem we have the following dataset:

18, 26, 19, 25, 19, 23, 31, 22, 19, 18, 20

The first step is sort the values on increasing order and we got:

18, 18, 19, 19, 19, 20, 22, 23. 25, 26, 31

And in order to find the first quartile we can separate the first 5 lowest values from the largest 5 values. And for this case we focus on these values:

18, 18, 19, 19, 19

And if we find the median of the last 5 values we can find the first quartile and we got:

$Q_1 = 19$

The best answer would be:

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4 years ago

What does the word "of" mean? <br>A. in addition to<br>B. multiply by<br>C. divided by

Natali [406]

Answer:

B. multiply by

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3 years ago

Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b

Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

Step-by-step explanation:

We are given the following in the question:

$p(x) = 0.1 e^{-0.1x}$

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

$=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%$

b) P(calls last 4 minutes or more)

$=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%$

c) cumulative distribution function

$P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b$

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4 years ago

I need help with this problem! I'll give brainlist! ​

I need help with this problem! I'll give brainlist!