Find the product of 0.34 and 0.02

Mathematics

2 answers:

Sedbober [7]3 years ago

7 0

It’s 0.68 good luck, I’m here for help

dmitriy555 [2]3 years ago

6 0

Answer:

<h2>The product is 0.68</h2>

Step-by-step explanation:

A product means we are multiplying. we are multiplying 0.34 and 0.02 which is 0.68

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This is 70 points I will report anyone who answers wrong just for points! I will be asking TWO questions Please answer correctly

Ratling [72]

QUESTION 1: In these type of question, the easiest way to get the answer is try to plug in the x and y values from the options given in the equation given, So in the first question all the choice except C are more then 14 if you plug in x and y's, for eg, if you plug in x = 3 and y = 2 , you get (3+3)2 = 14 6 x 2 = 14 12 is not equal to 14, so this eliminates this choice but if you chose C you get, (11+3)1 = 14 14 = 14 so this makes C the solution for first question and for the second question do the same thing, and the answer will be D. Hope this helps

QUESTION 2: 5xy + 9 = 44

5xy = 35

xy = 7

solution pairs are:

C. (1, 7) and (7, 1)

not mentioned: (-1.-7) and (-7, -1)

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Plzz don't forget to rate and thanks me

8 0

4 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