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Lelu [443]
3 years ago
13

The radius of Earth is about 6.38 x 10³ kilometers. Write this distance in standard notation.​

Mathematics
1 answer:
Nina [5.8K]3 years ago
7 0

Answer:

6380

Step-by-step explanation:

You can use your calculator or knowing that 10^3 is 1000 so 6.38 x 1000= 6380.

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What is the solution to the system of equations?
Minchanka [31]

Answer:

  (1, –5)

Step-by-step explanation:

It is relatively easy to try the offered solutions to see what works.

<u>(1, -5)</u>

  3(1) +10(-5) = -47 . . . true

  5(1) -7(-5) = 40 . . . true

(1, -5) is the solution

_____

As a check, you can try some of the other choices:

<u>(1, 5)</u>

  3(1) +10(5) ≠ -47

<u>(-1, -5)</u>

  3(-1) +10(-5) ≠ -47

<u>(-1, 5)</u>

  3(-1) +10(5) ≠ -47

None of the other choices works in the first equation, so they're not the solution.

4 0
3 years ago
What is the answer to 7(5-8x) &gt; 147
Liono4ka [1.6K]

Answer: x < -2

Step-by-step explanation:

7 0
3 years ago
Mileage tests are conducted for a particular model of automobile. If a 98% confidence interval with a margin of error of 1 mile
Llana [10]

Answer: 37

Step-by-step explanation:

Given : Significance level : \alpha:1-0.98=0.02

Critical value : z_{\alpha/2}=z_{0.01}=2.33

margin of error : E= 1 mile per gallon

Population standard deviation: \sigma=2.6 miles per gallon.

We know that when the population standard deviation is known then the formula to find the sample size is given by  :

n=(\dfrac{z_{\alpha/2}\times \sigma}{E})^2

n=(\dfrac{(2.33)\times 2.6}{1})^2=36.699364\approx37  [Round to the next whole number.]

Hence, the required number of automobiles should be used in the test = 37

7 0
3 years ago
Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

3 0
3 years ago
Find the area of a sector with a central angle of 125 degrees and a radius of 11cm. Express the approximate answer to the neares
White raven [17]
The answer is c I’m pretty aure
5 0
3 years ago
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