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2 years ago

0.92 is 10 times as much as

1 answer:

5 0

Well just divide and .92 divided by 10 = .092 because you move the value place. so the tenths go to the hundredths and get .092. .092 is the answer

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Simplify the expression by combining like terms. 16+8a−3a+6b−9

Svetradugi [14.3K]

7+5a+6b is the answer because 16-9 is 7 and 8-3 is 5.

4 0

3 years ago

Read 2 more answers

Is sqrt(25a^2/4) the same as sqrt(25a^2)/sqrt(4)?

Evgesh-ka [11]

$LHS\\ \\ =\sqrt { \frac { 25{ a }^{ 2 } }{ 4 } } \\ \\ ={ \left( \frac { 25{ a }^{ 2 } }{ 4 } \right) }^{ \frac { 1 }{ 2 } }$

$\\ \\ =\frac { { \left( 25{ a }^{ 2 } \right) }^{ \frac { 1 }{ 2 } } }{ { 4 }^{ \frac { 1 }{ 2 } } } \\ \\ =\frac { \sqrt { 25{ a }^{ 2 } } }{ \sqrt { 4 } } \\ \\ =RHS$

6 0

3 years ago

Read 2 more answers

Marvin scored 96, 93, 97, and 92 on four out of five math tests. In order to receive the highest grade in the class, he must obt

castortr0y [4]

Answer:98

Step-by-step explanation:

has to be higher then 97 so it has to be 98

5 0

2 years ago

Read 2 more answers

HELP WITH ALL PLZ I NEED HELP

alex41 [277]

1.) 16/100 and 0.16
2.) 7/10 and 0.7
3.) 6/10
4.) 73/100
5.) 6 9/10
6.)8 57/100
7.)0.70
8.)0.33
9.)7.20
10.)3.09
11.) 0.80
12.)0.48
13.)0.02
14.)0.55
15.)D
I hope all of these are correct and help

4 0

2 years ago

Assume that a simple random sample has been selected from a normally distributed population. State the hypotheses, find the test

BaLLatris [955]

Solution

Hypotheses:

- The population mean is 132. In order to test the claim that the mean is 132, we should check for if the mean is not 132.

- Thus, the Hypotheses are:

$\begin{gathered} H_0:\mu=132 \\ H_1:\mu\ne132 \end{gathered}$

Test statistic:

- The test statistic has to be a t-statistic because the sample size (n) is less than 30.

- The formula for finding the t-statistic is:

$\begin{gathered} t=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}} \\ \\ where, \\ \bar{X}=\text{ Sample mean} \\ \mu=\text{ Population mean} \\ s=\text{ Standard deviation} \\ n=\text{ Sample size} \end{gathered}$

- Applying the formula, we have:

$\begin{gathered} t=\frac{137-132}{\frac{14.2}{\sqrt{20}}} \\ \\ t=\frac{5}{3.1752} \\ \\ t\approx1.5747 \end{gathered}$

Critical value:

- The critical value t-critical, is gotten by reading off the t-distribution table.

- For this, we need the degrees of freedom (df) which is gotten by the formula:

$\begin{gathered} df=n-1 \\ df=20-1=19 \end{gathered}$

- And then we also use the significance level of 0.1 and the fact that it is a two-tailed test to trace out the t-critical. (Note: significance level of 0.1 implies 10% significance level)

- This is done below:

- The critical value is 1.729

P-value:

- To find the p-value, we simply check the table for where the t-statistic falls.

- The t-statistic given is 1.5747. We simply check which values this falls between in the t-distribution table. It falls between 1.328 and 1.729. We can simply choose a value between 0.1 and 0.05 and multiply the result by 2 since it is a two-tailed test.

- However, we can also use a t-distribution calculator, we have:

- Thus, the p-value is 0.13183

Final Conclusion:

- The p-value is 0.13183, and comparing this to the significance level of 0.1, we can see that 0.13183 is outside the rejection region.

- Thus, the result is not significant and we fail to reject the null hypothesis

7 0

1 year ago