There are 4 blue beads and 20 red beads

Add the following decimals 0.654+0.45

anzhelika [568]

When adding decimals you would line up the decimal points and fill blank spots with zeros.

so...

0.654

+ 0.450

⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻⁻

<u>1.104</u>.

6 0

3 years ago

Read 2 more answers

5 to the power 3 is equal to 125 Express in long automatic form

levacccp [35]

Answer is maybe 5x5x5=125

8 0

3 years ago

Let f(x) = |x| for all real numbers x. Write the formula for the function represented by the described

jeyben [28]

Answer:

y = (-3f(x+4))+5

Step-by-step explanation:

Please see the picture below.

1. Apply to the function the vertical stretch with scale factor 3:

y = 3f(x)

2. Multiply the function by -1 to reflect it over the x-axis:

y = -3f(x)

3. Add up 4 units from x to translate the function 4 units to the left:

y = -3f(x+4)

4. Add up 5 units to the entire function to translate it up 5 units:

y = (-3f(x+4))+5

3 0

2 years ago

Read 2 more answers

A university claims that the average cost of books per student, per semester is $300. A group of students believes that the actu

LenKa [72]

Answer:

$t=\frac{345-300}{\frac{200}{\sqrt{100}}}=2.25$

$p_v =P(t_{(99)}>2.25)=0.0133$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean is higher than 300 at 1% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=345$ represent the sample mean

$s=200$ represent the sample standard deviation

$n=100$ sample size

$\mu_o =300$ represent the value that we want to test

$\alpha$ 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 300, the system of hypothesis would be:

Null hypothesis: $\mu \leq 300$

Alternative hypothesis: $\mu > 300$

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:

$t=\frac{345-300}{\frac{200}{\sqrt{100}}}=2.25$

P-value

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

$df=n-1=100-1=99$

Since is a one right tailed test the p value would be:

$p_v =P(t_{(99)}>2.25)=0.0133$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the mean is higher than 300 at 1% of significance.

7 0

3 years ago

Multiply. 12ab(a^2+20ab-4)

ella [17]

= 12a^3b + 240a^2b^2 -48ab

4 0

3 years ago