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3 years ago

What term generally describes small chunks of rocks and debris in space? meteorites meteoroids meteors meteorides

Mathematics

2 answers:

Rus_ich [418]3 years ago

7 0

Answer:

When in space it is a meteorite. When it is entering the atmosphere, it is a meteor. When it hits the earth it;s classified as a meteoroid.

Hope this helps.

Step-by-step explanation:

deff fn [24]3 years ago

5 0

Answer:

meteoroids

Step-by-step explanation:

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What is 4^12 x 125^7? Express your answer in scientific notation

lianna [129]

Answer:

8×10^21

Step-by-step explanation:

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3 years ago

Function g is a transformation of the parent exponential function. Which statements are true about function g?

Elan Coil [88]

Answer:

function g is positive over (-∞, ∞)

function g has a y-intercept of (0,4)

function g decreasing over the interval (-∞, 0)

Step-by-step explanation:

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2 years ago

A simple random sample of size nequals81 is obtained from a population with mu equals 83 and sigma equals 27. (a) Describe the

Ivanshal [37]

Answer:

a) $\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

b) $z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

c) $z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

d) $z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

Step-by-step explanation:

For this case we know the following propoertis for the random variable X

$\mu = 83, \sigma = 27$

We select a sample size of n = 81

Part a

Since the sample size is large enough we can use the central limit distribution and the distribution for the sampel mean on this case would be:

$\bar X \sim N (\mu, \frac{\sigma}{\sqrt{n}})$

With:

$\mu_{\bar X}= 83$

$\sigma_{\bar X}=\frac{27}{\sqrt{81}}= 3$

Part b

We want this probability:

$P(\bar X>89)$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 89 we got:

$z= \frac{89-83}{\frac{27}{\sqrt{81}}}= 2$

$P(Z>2) = 1-P(Z$

Part c

$P(\bar X$

We can use the z score formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for 75.65 we got:

$z= \frac{75.65-83}{\frac{27}{\sqrt{81}}}= -2.45$

$P(Z$

Part d

We want this probability:

$P(79.4 < \bar X < 89.3)$

We find the z scores:

$z= \frac{89.3-83}{\frac{27}{\sqrt{81}}}= 2.1$

$z= \frac{79.4-83}{\frac{27}{\sqrt{81}}}= -1.2$

$P(-1.2$

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3 years ago

2 3/4÷2/5 will mark brainleyst

olga55 [171]

Answer:

6 7/8

Step-by-step explanation:

Result in decimals: 6.875

In pic

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2 years ago

Read 2 more answers

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Mademuasel [1]

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Step-by-step explanation: this will help you out

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2 years ago