All categories

3 years ago

I need help. thanks? lolz.

Mathematics

1 answer:

Tpy6a [65]3 years ago

3 0

The answer is the second one aka 6

You might be interested in

Assume that 0 a. 16/63 b. -16/63 c. 63/16 d. -63/16 please help

iVinArrow [24]

The answer to this question is B

3 0

3 years ago

Use the interactive tool to find which of the following expressions have a sum of 6. Check all that apply.

Gwar [14]

Answer:

(–3) + 9 , 7 + (–1)

Step-by-step explanation:

srry im late and srry if im wrong

8 0

3 years ago

110 ft below sea level positive or negative

ryzh [129]

Negative because it is below sea level.

8 0

3 years ago

Read 2 more answers

The manager at the local auto shop has found that the probability that a car brought into the shop requires an oil change is 0.8

Ber [7]

Answer:

0.85

Step-by-step explanation:

Given two events A and B, the probability that either A or B occurs is given by:

$p(A\cup B) = p(A)+p(B)-p(A\cap B)$

where

$p(A)$ the probability that A occurs

$p(B)$ is the probability that B occurs

$p(A\cap B)$ is the probability that both A and B occur at the same time

In this problem, we know the following facts:

$p(o) = 0.83$ is the probability that the car requires an oil change

$p(b)=0.17$ is the probability that the car requires a brake repair

$p(o\cap b) = 0.15$ is the probability that the car requires both an oil change and brake repair

Therefore, the probability that either o (car requiring oil change) or b (car requiring brake repait) occur is:

$p(o\cup b)=p(o)+p(b)-p(o\cap b)=0.83+0.17-0.15 = 0.85$

4 0

3 years ago

A statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20. Use thi

rjkz [21]

Answer:

\mu = 14.5\\

\sigma = 5.071\\

k = 1.084

Step-by-step explanation:

given that a statistician uses Chebyshev's Theorem to estimate that at least 15 % of a population lies between the values 9 and 20.

i.e. his findings with respect to probability are

$P(9$

Recall Chebyshev's inequality that

$P(|X-\mu |\geq k\sigma )\leq {\frac {1}{k^{2}}}\\P(|X-\mu |\leq k\sigma )\geq 1-{\frac {1}{k^{2}}}\\$

Comparing with the Ii equation which is appropriate here we find that

$\mu =14.5$

Next what we find is

$k\sigma = 5.5\\1-\frac{1}{k^2} =0.15\\\frac{1}{k^2}=0.85\\k=1.084\\1.084 (\sigma) = 5.5\\\sigma = 5.071$

Thus from the given information we find that

$\mu = 14.5\\\sigma = 5.071\\k = 1.084$

5 0

3 years ago