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1 year ago

15

Rohan had ten dollars. he bought x pounds of potatoes which were on sale for 40% off the original price. write and expression th

at represents the amount of money rohan had left in simplest terms.

1 answer:

yanalaym [24]1 year ago

5 0

Let y represent the original price of potatoes per pound

Since he purchased the potatoes at 40% off the original price, this can represented as y*(1-0.4)

Now the weight of potatoes he purchased is x pounds, the total cost of these potatoes can be represented as x*y*(1-0.4)

The total amount of money he had is $10

The amount of money left over can be represented by the expression =

10 - x*y*(1-0.4)

where "x" is the weight of potatoes in pounds and "y" is the original price per pound of potatoes

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The graph represents function 1, and the equation represents function 2: A coordinate plane graph is shown. A horizontal line is

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1 year ago

Annie is running a trail that is 4 1/2 miles long. She ran the first 2/3 of the distance. Then the trail became very steep, so s

qwelly [4]

Answer:

Annie walked 1 11/12 miles

Step-by-step explanation:

Trail= 4 1/2 miles

Annie ran the first 2/3 miles

Remaining miles= Total miles - Miles covered first

= 4 1/2 - 2/3

= 9/2 - 2/3

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1 year ago

An investor believes that investing in domestic and international stocks will give a difference in the mean rate of return. They

arlik [135]

Answer:

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X_1 =2.0233$ represent the sample mean 1

$\bar X_2 =3.048$ represent the sample mean 2

n1=15 represent the sample 1 size

n2=15 represent the sample 2 size

$s_1 =4.893387$ population sample deviation for sample 1

$s_2 =5.12399$ population sample deviation for sample 2

$\mu_1 -\mu_2$ parameter of interest at 0.1 of significance so the confidence would be 0.9 or 90%

We want to test:

H0: $\mu_1 = \mu_2$

H1: $\mu_1 \neq \mu_2$

And we can do this using the confidence interval for the difference of means.

Solution to the problem

The confidence interval for the difference of means is given by the following formula:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}$ (1)

The point of estimate for $\mu_1 -\mu_2$ is just given by:

$\bar X_1 -\bar X_2 =2.0233-3.048=-1.0247$

The degrees of freedom are given by:

$df = n_1 +n_2 -2 = 15+15-2=28$

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 value. The excel command would be: "=-T.INV(0.05,28)".And we see that $t_{\alpha/2}=\pm 1.701$

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

$-1.0247-1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=-4.137$

$-1.0247+1.701\sqrt{\frac{4.893387^2}{15}+\frac{5.12399^2}{15}}=2.087$

So on this case the 90% confidence interval would be given by $-4.137 \leq \mu_1 -\mu_2 \leq 2.087$

For this case since the confidence interval for the difference of means contains the 0 we can conclude that we don't have significant differences at 10% of significance between the two means analyzed.

4 0

1 year ago

A jet flies 425 km from Ottawa to Québec at rate v + 60. On the return flight, the

Marina CMI [18]

Answer:

a. $\frac{- 42,500}{(v + 60)(v - 40)}$

Step-by-step Explanation:

Given:

Distance Ottawa to Québec = 425 km

Initial flight rate = v + 60

Return flight rate = v - 40

$t = \frac{d}{r}$

Required:

Flight times difference of the initial and return flights

Solution:

=>Flight time of the initial flight:

$t = \frac{d}{r}$

$t = \frac{425}{v + 60}$

=>Flight time of the return flight:

$t = \frac{425}{v - 40}$

=>Difference in flight times:

$\frac{425}{v + 60} - \frac{425}{v - 40}$

$\frac{425(v - 40) -425(v + 60)}{(v + 60)(v - 40)}$

$\frac{425(v) - 425(40) -425(v) -425(+60)}{(v + 60)(v - 40)}$

$\frac{425v - 17000 -425v - 25500}{(v + 60)(v - 40)}$

$\frac{425v - 425v - 17000 - 25500}{(v + 60)(v - 40)}$

$\frac{- 42,500}{(v + 60)(v - 40)}$

3 0

2 years ago