All categories

3 years ago

Simplify 4! A. 24 B. 10 C. 9 D. 4

Mathematics

1 answer:

professor190 [17]3 years ago

5 0

Answer:

Step-by-step explanation:

You might be interested in

PLSS ANSWER MY QUESTION

Taya2010 [7]

Answer:

1) Area of frame: $140cm^2$ ; Area of picture: $40cm^2$ ; Total Area: $180cm^2$ .

2) Area 1 = $48cm^2$ ; Area 2 = $12cm^2$ ; total area = $60cm^2$ [first answer line : $60cm^2$ ]

Step-by-step explanation:

1. So to answer this question, you have to find the inner photo area first (which you will then subtract from the overall rectangle area to get the photo frame area).

To find area, you multiply length x width. For this question, there are not enough measurements given to initially multiply length and width. So, assuming that the photo is centered on the frame and the frame width is consistent along all sides, you find the difference between the overall length (12) and the picture length (5).

12 - 5 = 7

The leftover area (outside of the picture) is 7 cm (half of seven on both sides), so one side can be assumed to be : 3.5 cm

this means that 15 cm - 3.5 cm of both sides lengths (difference between the two lengths given) is the height of the photo, 15 - 3.5(2) = 8

So, the dimensions/measurements of the photo are 5 cm by 8 cm.

This means that the inner photo area is 40 (5 x 8) cm

The area of the frame will be the area of the total rectangle [12 x 15] - the inner photo area [40]. [ (12)(15) - 40 ]

[180 - 40] = 140 cm

(area is measured in squared units, so the area of the frame should be written as $140cm^{2}$ ; and the area of the photo should be written as $40cm^{2}$ )

The total area is the entire area of the rectangle: (12)(15) = 180. (This should be written as $180cm^2$ )

--------------------------------------

2) Finding the area of a shape that isn't a simple rectangle is easier if you separate the shape into smaller rectangles. I will separate the two rectangles into 4 x 12 and 2 x 6 [6 cm would be the remaining side length if you remove the length of 4].

{you can separate the shape in multiple ways, so my answer is only one answer to this problem out of other splits.}

as mentioned, the area is length x width.

4 x 12 = 48.

2 x 6 = 12. Once again, you should write area in squared units.

(the area 1 should be written as $48cm^{2}$ ; the area 2 should be written as $12cm^2$ )

you can combine the separated areas (add them) to find the total area. 48 + 12 = 60.

(the total area should be written in squared units also, so the total area should be written as: $60cm^2$ )

The area of the figure is $60cm^2$

8 0

2 years ago

I really need this answered so I can understand how to do the rest.

qwelly [4]

Answer:

Step-by-step explanation:

We can write

$4=\sqrt{\frac{3v}{3.14*8}}\\4^2=\frac{3v}{25.12}\\16*25.12=3v\\401.92=3v\\133.9733333333333=v$

4 0

2 years ago

An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co

levacccp [35]

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)

2. $4.7048 \leq \sigma^2 \leq 16.1961$

3. $t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

$t_{crit}=1.753$

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. $t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

$\chi^2 =24.9958$

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

$\bar X=26.31$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=16-1=15$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that $t_{\alpha/2}=2.13$

Now we have everything in order to replace into formula (1):

$26.31-2.13\frac{2.8}{\sqrt{16}}=24.819$

$26.31+2.13\frac{2.8}{\sqrt{16}}=27.801$

So on this case the 95% confidence interval would be given by (24.8190;27.8010)

Part 2

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=16-1=15$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical values.

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

$\chi^2_{\alpha/2}=24.996$

$\chi^2_{1- \alpha/2}=7.261$

And replacing into the formula for the interval we got:

$\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}$

$4.7048 \leq \sigma^2 \leq 16.1961$

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:

Null hypothesis: $\mu \leq 25$

Alternative hypothesis: $\mu > 25$

We don't know the population deviation, so for this case 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)

Calculate the statistic

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

$t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871$

Critical value

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got $t_{crit}=1.753$

Conclusion

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: $\sigma \leq 2.3$

H1: $\sigma >2.3$

In order to check the hypothesis we need to calculate the statistic given by the following formula:

$t=(n-1) [\frac{s}{\sigma_o}]^2$

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

$t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306$

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be $\chi^2 =24.9958$

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

7 0

3 years ago

What is the probability that you would pick a blue marble and roll a 5 OR 6?

NNADVOKAT [17]

Marble: 2/7

Dice: 2/6 = 1/3

7 0

2 years ago

Read 2 more answers

X^3+6x^2-17x+2-x^3-x^2-11x+36

Rudiy27

<span>X^3+6x^2-17x+2-x^3-x^2-11x+36 = </span>5x^2 - 28x + 38

4 0

3 years ago