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postnew [5]
3 years ago
12

A truck carries apples, grapes, and blackberries in the ratio of 4:3:4 if the apples weigh 160 pounds, how much the the truckloa

d of fruit weigh in total
Mathematics
1 answer:
pishuonlain [190]3 years ago
4 0

Answer:the truck load of fruit weigh 440 pounces

Step-by-step explanation:

Let the total weight of the truck load of fruit weigh x pounds.

The truck carries apples, grapes, and blackberries in the ratio of 4:3:4

The total ratio is the sum of the proportions of apples, grapes, and blackberries. It becomes 4+3+4 = 11

if the apples weigh 160 pounds, it means that

4/11 × x = 160

4x/11 = 160

4x = 160×11 = 1760

x = 1760/4

x = 440

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C=3 \times 2500 + 425=7500+425=7925

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3 years ago
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what is the original price and the markup to find the retail price. Original price: $44; Markup: 7.5%
torisob [31]

Answer:

the retail price is $47.30

Step-by-step explanation:

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2 years ago
Odysseus found that his troubles varied directly as his distance from his home island of Ithaca. If he had 20 troubles when he w
Vitek1552 [10]
Lol


trouble varies directly as distance
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2 years ago
Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a
tigry1 [53]

Answer:

SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58

ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073

\bar X - ME = 237- 14.073 = 222.927

\bar X + ME = 237+ 14.073 = 251.073

n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis:\mu \geq 247  

Alternative hypothesis:\mu  

z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165  

p_v =P(z  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

\bar X =237 represent the sample mean

\sigma = 47 represent the population deviation

n =30 represent the sample size selected

\mu_0 = 247 represent the value that we want to test.

The standard error for this case is given by:

SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58

For the 90% confidence the value of the significance is given by \alpha=1-0.9 = 0.1 and \alpha/2 = 0.05 so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

z_{\alpha/2}= 1.64

And the margin of error would be:

ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073

The confidence interval for this case would be given by:

\bar X - ME = 237- 14.073 = 222.927

\bar X + ME = 237+ 14.073 = 251.073

The margin of error is given by this formula:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}    (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

n=(\frac{z_{\alpha/2} \sigma}{ME})^2   (b)

Replacing into formula (b) we got:

n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:  

Null hypothesis:\mu \geq 247  

Alternative hypothesis:\mu  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-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:  

z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165  

P-value  

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

p_v =P(z  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

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Effectus [21]

1/2 or 0.5 hope this helps

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