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3 years ago

What is each number in scientific notation? Drag the answer into the box to match each number. 0.0000855 ?8.55×10?5? 855,000

Mathematics

2 answers:

Vera_Pavlovna [14]3 years ago

7 0

I'm confused on what you're asking, but this is scientific notation on the first one of 0.0000855. 8.55 x 10^-5

mel-nik [20]3 years ago

5 0

Answer:

$8.55\times 10^{-5}$

Step-by-step explanation:

Given number,

0.0000855,

Since, scientific notation is way to write a large or small number as the product of a number between 0 to 9 and an exponent of 10,

Also, when we move the decimal towards right side to obtain the number between 0 to 9 the exponent of 10 is negative value of shifting otherwise it is positive value of shifting,

Thus, by the above explanation,

The scientific notation of the number 0.0000855 is $8.55\times 10^{-5}$

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Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 signif

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Answer:

a) F. H 0 : μ O ≥ μ C

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Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

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$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis: $\mu_{O} \geq \mu_{C}$

Alternative hypothesis: $\mu_{O} < \mu_{C}$

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$

Part f

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Answer:

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2 years ago