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3 years ago

What is 4,700,930,002 in expanded form

Mathematics

2 answers:

Svetradugi [14.3K]3 years ago

5 0

4,000,000,000 + 700,000,000 + 900,000 + 30,000 + 2 is 4,700,930,002 in expanded form!

I hope this helps! :>

soldier1979 [14.2K]3 years ago

3 0

4,000,000,000 + 700,000,000 + 0 + 0 + 900,000 + 30,000 + 0 + 0 + 0 + 2

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. (07.01) Karolyn is arranging her art project, science project, and math project on her desk. The tree diagram below shows the

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The tree diagram shows that there are a total of six sample spaces.

math------science

art

science---math

art---math

science

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math

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Based on the tree diagram, the probability that the science project is first is two over six. This is because science appears first in two out of the six sample spaces (boldened).

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3 years ago

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The life in hours of a biomedical device under development in the laboratory is known to be approximately normally distributed.

tatuchka [14]

Answer:

a)Null hypothesis:- H₀: μ> 500

Alternative hypothesis:-H₁ : μ< 500

b) (5211.05 , 5411.7)

95% lower confidence bound on the mean.

c) The test of hypothesis t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Step-by-step explanation:

Step :-1

Given a random sample of 15 devices is selected in the laboratory.

size of the small sample 'n' = 15

An average life of 5311.4 hours and a sample standard deviation of 220.7 hours.

Average of sample mean (x⁻) = 5311.4 hours

sample standard deviation (S) = 220.7 hours.

Step :- 2

a) Null hypothesis:- H₀: μ> 500

Alternative hypothesis:-H₁ : μ< 500

Level of significance :- α = 0.95 or 0.05

b) The test statistic

$t = \frac{x^{-} - mean}{\frac{S}{\sqrt{n-1} } }$

$t = \frac{5311.4 - 500}{\frac{220.7}{\sqrt{15-1} } }$

t = 5.826

The degrees of freedom γ= n-1 = 15-1 =14

tabulated value t =1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

calculated value t = 5.826 >1.761 From 't' distribution table at 14 degrees of freedom at 95% level of significance.

Null hypothesis is rejected at 95% confidence on the mean.

C) The 95% of confidence limits

$(x^{-} - t_{0.05} \frac{S}{\sqrt{n} } ,x^{-} + t_{0.05}\frac{S}{\sqrt{n} } )$

substitute values and simplification , we get

$(5311.4 - 1.761 \frac{220.7}{\sqrt{15} } ,5311.4 +1.761\frac{220.7}{\sqrt{15} } )$

(5211.05 , 5411.7)

95% lower confidence bound on the mean.

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3 years ago

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The answer is 45 okay bro good

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2 years ago

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Arte-miy333 [17]

Your Answer: A.) (2 , 5) I also provided the steps.

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To test Upper H 0: muequals50 versus Upper H 1: muless than50, a random sample of size nequals23 is obtained from a populatio

natta225 [31]

Answer:

Step-by-step explanation:

Hello!

To test H0: u= 50 versus H1= u < 50, a random sample size of n = 23 is obtained from a population that is known to be normally distributed. Complete parts A through D. 

 A) If ¯ x = 47.9 and s=11.9, compute the test statistic .

For thistest the corresponsing statistis is a one sample t-test

$t= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }~~t_{n-1}$

$t_{H_0}= \frac{47.9-50}{\frac{11.9}{\sqrt{23} } } = -0.846= -0.85$

B) If the researcher decides to test this hypothesis at the a=0.1 level of significance, determine the critical value(s).

This test is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of the statistic. The ejection region is defined by one critical value:

$t_{n-1;\alpha }= t_{22;0.1}= -1.321$

Check the second attachment. The first row shows α= Level of significance; the First column shows ν= sample size.

The t-table shows the values of the statistic for the right tail. P(tₙ≥α)

But keep in mind that this distribution is centered in zero, meaning that the right and left tails are numerically equal, only the sign changes. Since in this example the rejection region is one-tailed to the left, the critical value is negative.

C) What does the distribution graph appear like?

Attachment.

D) Will the researcher reject the null hypothesis?

As said, the rejection region is one-tailed to the right, so the decision rule is:

If $t_{H_0}$ ≤ -1.321, reject the null hypothesis.

If $t_{H_0}$ > -1.321, do not reject the null hypothesis.

$t_{H_0}$ = -0.85, the decision is to not reject the null hypothesis.

To test H0: μ=100 versus H1:≠100, a simple random sample size of nequals=24 is obtained from a population that is known to be normally distributed. Answer parts (a)-(d).

a) If x =104.2 and s=9.6, compute the test statistic.

For this example you have to use a one sample t-test too. The formula of the statistic is the same:

$t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } = \frac{104.2-100}{\frac{9.6}{\sqrt{24} } = } = 2.143$

b) If the researcher decides to test this hypothesis at the α=0.01 level of significance, determine the critical values.

This hypothesis pair leads to a two-tailed rejection region, meaning, you'll reject the null hypothesis at either small or big values of the statistic. Then the rejection region is divided into two and determined by two critical values (the left one will be negative and the right one will be positive but the module of both values will be equal).

$t_{n-1;\alpha/2 }= t_{23; 0.005}= -2.807$

$t_{n-1;1-\alpha /2}= t_{23;0.995}= 2.807$

c) Draw a t-distribution that depicts the critical region(s). Which of the following graphs shows the critical region(s) in thet-distribution?

Attachment.

(d) Will the researcher reject the null hypothesis?

The decision rule for the two-tailed hypotheses pair is:

If $t_{H_0}$ ≤ -2.807 or if $t_{H_0}$ ≥ 2.807, reject the null hypothesis.

If -2.807 < $t_{H_0}$ < 2.807, do not reject the null hypothesis.

$t_{H_0}$ = 2.143 is greater than the right critical value, the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the test statistic is not between the critical values.

Full text in attachment. The sample size is different by 2 but it should serve as a good example.

H₀: μ = 20

H₁: μ < 20

a) n= 18, X[bar]= 18.3, S= 4, Compute statistic.

$t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }= \frac{18.3-20}{\frac{4}{\sqrt{18} } } = -1.80$

b) The rejection region in this example is one-tailed to the left, meaning that you'll reject the null hypothesis to small values of t.

Out of the three graphics, the correct one is A.

To resolve this you have to look for the values in the t-table that are the closest to the calculated $t_{H_0}$

Symbolically:

$t_{n-1;\alpha_1 } \leq t_{H_0}\leq t_{n-1;\alpha _2}$

$t_{H_0}$ = -1.80

$t_{17; 0.025 }= -2.110$

$t_{17;0.05}= -1.740$

Roughly defined you can say that the p-value is the probability of obtaining the value of $t_{H_0}$ , symbolically: P(t₁₇≤-1.80)

Under the distribution the calculated statistic is between the values of -2.110 and -1.740, then the p-value will be between their cumulated probabilities:

A. 0.025 < p-value < 0.05

d. The researcher decides to test the hypothesis using a significance level of α: 0.05

Using the p-value approach the decision rule is the following:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

We already established in item c) that the p-value is less than 0.05, so the decision is to reject the null hypothesis.

Correct option:

B. The researcher will reject the null hypothesis since the p-value is less than α.

I hope this helps!

6 0

3 years ago