Ask question

Ask question

All categories

Alenkasestr [34]

3 years ago

13

Type the correct answer in each box. Use numerals instead of words. If necessary, use/ for the fraction bar(s).Find the inverse

of the given function.

2 answers:

artcher [175]3 years ago

8 0

Answer:

5,7,8

Step-by-step explanation:

Idk I’m not sure lemme know

Soloha48 [4]3 years ago

7 0

Answer:

First Blank: -7

Second Blank: 1

Third Blank: 5

Fourth Blank: -5

Step-by-step explanation:

You might be interested in

Plzz Help ASAP!! i posted the picture?

Andre45 [30]

Hi!

So the original formula for a problem likes this or a quadratic is:

ax^2 + bx+ c

so a is the accelaration; b is the initial velocity; and c is the initial height

so none of those are the zeros, which means c is out

zeros are the x values when y is 0, that also means when the parabola crosses the x axis

so d is out too because the maximum height is the vertex which mean that there would probably be a y value. in some cases the vertex is the zero, but not here

B is out because x isn't time it is distance

your answer is A!

Hope this helps!

8 0

3 years ago

Read 2 more answers

Help quickly please!!!

nadezda [96]

The value of the expression is 12

6 0

4 years ago

8751 million as a number

Galina-37 [17]

8,751,000,000.
......

6 0

3 years ago

A researcher believes that the mean weight of competitive runners is about 140 pounds. A sample of 24 elite distance runners has

ExtremeBDS [4]

Answer:

$t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781$

$p_v =P(t_{(23)}$

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.

Step-by-step explanation:

Data given and notation

$\bar X=136$ represent the sample mean

$s=11$ represent the sample standard deviation

$n=24$ sample size

$\mu_o =140$ represent the value that we want to test

$\alpha=0.01$ 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 less than 140, the system of hypothesis would be:

Null hypothesis: $\mu \geq 140$

Alternative hypothesis: $\mu < 140$

If we analyze the size for the sample is < 30 and 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:

$t=\frac{136-140}{\frac{11}{\sqrt{24}}}=-1.781$

P-value

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

$df=n-1=24-1=23$

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

$p_v =P(t_{(23)}$

Conclusion

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the mean is less than 140 pounds at 5% of signficance.

4 0

3 years ago

A rectangular piece of land borders a wall. The land is to be enclosed and to be into divided 3 equal plots with 200 feet of fen

baherus [9]

Answer:

Area = 2500 square feet is the largest area enclosed

Step-by-step explanation:

A rectangular piece of land borders a wall. The land is to be enclosed and to be into divided 3 equal plots with 200 feet of fencing

Let x be the length of each box and y be the width of the box

Perimeter of the box= 3(length ) + 4(width)

$200=3x+4y$

solve for y

$200=3x+4y$

$200-3x=4y$

divide both sides by 4

$y=50-\frac{3x}{4}$

Area of the rectangle = length times width

$Area = 3x \cdot y$

$Area = 3x \cdot (50-\frac{3x}{4})$

$A=150x-\frac{9x^2}{4}$

Now take derivative

$A'=150-\frac{9x}{2}$

Set it =0 and solve for x

$0=150-\frac{9x}{2}$

$150=\frac{9x}{2}$

multiply both sides by 2/9

$x=\frac{100}{3}$

$A''=-\frac{9}{2}$

For any value of x, second derivative is negative

So maximum at x= 100/3

$A=150x-\frac{9x^2}{4}$ , replace the value of x

$A=150(\frac{100}{3})-\frac{9(\frac{100}{3})^2}{4})$

Area = 2500 square feet is the largest area enclosed

3 0

3 years ago