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3 years ago

MARKING BRAINLY PLEASE JUST HELPPP

Mathematics

2 answers:

Keith_Richards [23]3 years ago

8 0

Answer:

i don't it is correct or not........

Step-by-step explanation:

The Mass of Earth = 5.9742 × 1024 kilograms == == The Earth is (As of May 2000) is thought to be approximately 5.972 sextillion (5,972,000,000,000,000,000,000) metric tons.

A metric ton is a thousand kilograms, or 2,204.62 pounds.

Approximately 6,000,000,000,000,000,000,000,000 (6x1024) kilograms.

mamaluj [8]3 years ago

7 0

Answer:

Mass on water on earth is 29.5×10^20 kg approximately.

Step-by-step explanation:

Mass of earth= 5.9726 × 10^ 24 kg

water's mass is only 5 × 10 -4 percent of the Earth's total mass.

so mass of water = 5.9726×10^24 kg×5×10^-4=29.8603×10^20kg approximately

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How many models oder 100 do you need to model 3200? Explain.

Nat2105 [25]

3200÷100=32
Since each model is 100, you need 32 to model 3,200

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3 years ago

Please answer with proof please!!!

qwelly [4]

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3 years ago

Read 2 more answers

Add 7 to me. Then subtract 10. If you subtract 5 and then add 9, you get 14. What number am I?

Vesna [10]

Answer:

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3 years ago

The sale price of every item in a store is 75% of its usual price. A. The usual price of a jacket is $80, what is its sale price

riadik2000 [5.3K]

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8 0

3 years ago

Persons having Raynaud's syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experime

Citrus2011 [14]

Answer:

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

Step-by-step explanation:

We have to perform a hypothesis test on the difference between means.

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2$

μ1: mean heat output for subjects with the syndrome.

μ2: mean heat output for non-sufferers.

We will use a significance level of 0.05.

The difference between sample means is:

$M_d=\bar x_1-\bar x_2=0.63-2.09=-1.46$

The standard error is

$s_{M_d}=\sqrt{\sigma_1^2/n_1+\sigma_2^2/n_2}=\sqrt{0.3^2/9+0.5^2/9}=\sqrt{ 0.038 } \\\\ s_{M_d}=0.194$

The t-statistic is

$t=\dfrac{M_d}{s_{M_d}}=\dfrac{-1.46}{0.194}=-7.52$

The degrees of freedom are

$df=n_1+n_2-2=9+9-2=16$

The critical value for a left tailed test at a significance level of 0.05 and 16 degrees of freedom is t=-1.746.

The t-statistic is below the critical value, so it lies in the rejection region.

The null hypothesis is rejected.

There is enough evidence to say that the true average heat output of persons with the syndrmoe differs from the true average heat output of non-sufferers.

5 0

3 years ago