Pleeeeeease help me soon

Find in the table for any missing values. Please help!!!

Oxana [17]

Answer:

1:3

2:6

3:9

5:15

Step-by-step explanation:

Ask yourself: What times 3 equals my number?

As you can see 1x3=3 then 2x3=6, 3x3=9, and lastly 5x3=15

This is how you solve the chart.

8 0

3 years ago

15 points!!<br> Look at the picture to see the problem. What do you think?

Anna007 [38]

Answer:

I don't remember well, and I'm not sure if this is correct, but, I think the answer is This is beacuse ∠FCD are facing with ∠CDG meaning it would be a same side interior angle.

6 0

2 years ago

Read 2 more answers

Does anyone know how to foil the answer ?!

Liono4ka [1.6K]

Answer:

(a+b)(c+d)=a c+a d+b c+b d is the formula

Step-by-step explanation:

7 0

3 years ago

Does anyone how to solve this sum? It’s urgent

Galina-37 [17]

The required value for the sum is 9580.

$\frac{10000}{(1+\frac{0.115}{4})^2}+68\frac{1-\frac{1}{(1+\frac{0.115}{4})^2}}{\frac{0.115}{4}}$

<h3>What is simplification?</h3>

Simplification in mathematics to solve the given condition on its operators.

$\frac{10000}{(1+\frac{0.115}{4})^2}+68\frac{1-\frac{1}{(1+\frac{0.115}{4})^2}}{\frac{0.115}{4}}$

= $\frac{10000}{1.028^2} +68*\frac{1-\frac{1}{1.028^2} }{\frac{0.115}{4} } \\9451+68*\frac{1-0.94}{\frac{0.115}{4} } \\\\9451+68*\frac{0.054}{\frac{0.115}{4} } \\\\\\9451+3.72{\frac{4}{0.115} } \\\\\\9451+129\\$

= 9580

The required solution is given as 9580.

Learn more about simplification here:
brainly.com/question/9218183

#SPJ1

4 0

2 years ago

A random sample of n = 40 observations from a quantitative population produced a mean x = 2.2 and a standard deviation s = 0.29.

zmey [24]

Answer:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_{(39)}>2.18)=0.0177$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X=2.2$ represent the sample mean

$s=0.29$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =2.1$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 2,1, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2.1$

Alternative hypothesis: $\mu > 2.1$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_{(39)}>2.18)=0.0177$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

7 0

3 years ago