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3 years ago

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Which expression is equivalent to the following complex fraction? StartFraction 2 Over x EndFraction minus StartFraction 4 Over

y EndFraction divided by StartFraction negative 5 Over y EndFraction + StartFraction 3 Over x EndFraction StartFraction 3 y + 5 x Over 2 (y minus 2 x) EndFraction StartFraction 2 (y minus 2 x) Over 3 y minus 5 x EndFraction StartFraction 2 (y minus 2 x) (3 y minus 5 x) Over x squared y squared EndFraction StartFraction x squared y squared Over 2 (y minus 2 x) (3 y minus 5 x) EndFraction

1 answer:

aliina [53]3 years ago

8 0

Answer:

its A

Step-by-step explanation:

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Find the amount of the discount on a $140 bicycle that is on sale for 30% off A.$98.00 B.$32.00 C.$42.00 D.$4.20

Marrrta [24]

140 * 0.30 = 42

the discount would be C. $42.00

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3 years ago

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A bicycle wheel with radius 30cm made 1000 revolutions. What distance did it cover?

Inga [223]

The distance covered by bicycle is 1.885 km.

What is distnace?

Distance is a measurement of how far apart two things or points are, either numerically or occasionally qualitatively. Distance can refer to a physical length in physics or to an estimate based on other factors in ordinary language.

Given: A bicycle wheel with radius 30cm made 1000 revolutions.

We have to find the total distance covered by bicycle.

First to find the circumference of the bicycle wheel.

Circumference = 2πr

where, r is the radius.

Here r = 30.

So,

Circumference = 2 x π x 30 = 60π

Now to find the total distance,

D = circumference x number of revolutions.

D = 60π x 1000

D = 188,495.5592 cm

D = 188,495.5592 /1000 km

D = 1.88 km

Therefore, the distance covered by bicycle is, 1.885 km.

To know more about the distance, click on the link

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1 year ago

Jennifer's saving to buy a bike. The bike costs $245. She has $125 saved, and each week she adds $15 to her savings. How long wi

Novosadov [1.4K]

It will take her 8 weeks to save up enough money for the bike.

To get the answer first you have to subtract 125 from 245 to see how much money she still has to save then divide the answer which is 120 by 15 and you get 8. Hope this helps!!

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3 years ago

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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

Find the area of that with explanations thanks

TEA [102]

Answer:

900m or 300m

Step-by-step explanation:

45 x 20 = 900

30 x 10 = 300

3 0

3 years ago

Read 2 more answers