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FinnZ [79.3K]
3 years ago
8

What's the answer to: divide £48 in the ratio 7:5 (please show your working, much appreciated)

Mathematics
1 answer:
Brums [2.3K]3 years ago
4 0
I can see you come from outside the USA. No insults intended!!
.
Anyways, we have to put one side of the ratio and match it to 48. 
.
Or even better...
.
MAKE!!
A!!
PROPORTION!!
.
So... Let's say 7 is proportional to 48... so 7/5 = 48/x, where x is the number equivalent to ratio of 5.
.
Cross multiply!!
<span>.
</span>7x = 240.
.
Divide 7 on both sides.
.
x = 34 2/7.
.
Hope I helped!!
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Your fuel expense for last year was $800 this year it was $960 what was the percentage of increase. A.23% B.18% C. 15% D.20%
zhannawk [14.2K]

Answer:

The percentage of increase in fuel expenses is 20%

Step-by-step explanation:

Given:

Fuel expense in last year =  $800

Fuel expense in this year =  $960

To Find:

The percentage of increase

Solution:

Let the percentage increase be X

Then X  =\frac{\text{final value}-\text{starting value}}{\text{starting value}}\times 100

Now substituting the values we get,

X =\frac{960-800}{800}\times 100

X =\frac{160{800}\times 100

X =0.2 \times 100

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Use a net to find the surface area of the cone to the nearest square centimeter. Use 3.14 for pi.
Lerok [7]

The surface area of the cone is given by:

SA=\pi r^2+\pi Lr

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In this case r=5 and L=12. Then the surface area is:

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In recent years, only 79% of the fruit sold at Fred's Fruit stand has been any good. Fred recently started buying fruit from new
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Answer:

The p-value of the test is 0.0485 < 0.05, also less than 0.07, so there is sufficient evidence to conclude that there is a statistically significant increase in the percentage of good fruit, for both significance levels, thus the change in significance levels do not change the conclusion.

Step-by-step explanation:

79% of the fruit sold at Fred's Fruit stand has been any good. Test if there has been an increase.

At the null hypothesis, we test if there has been no increase, that is, the proportion is still of 79%, so:

H_0: p = 0.79

At the alternative hypothesis, we test if there has been an increase, that is, more than 79% being good, so:

H_1: p > 0.79

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

0.79 is tested at the null hypothesis:

This means that \mu = 0.79, \sigma = \sqrt{0.79*0.21}

In a random sample if 475 pieces, 85 were bad.

So 475 - 85 = 390 were good, and:

n = 475, X = \frac{390}{475} = 0.8211

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{0.8211 - 0.79}{\frac{\sqrt{0.79*0.21}}{\sqrt{475}}}

z = 1.66

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion above 0.8211, which is 1 subtracted by the p-value of Z = 1.66.

Looking at the z-table, z = 1.66 has a p-value of 0.9515.

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The p-value of the test is 0.0485 < 0.05, also less than 0.07, so there is sufficient evidence to conclude that there is a statistically significant increase in the percentage of good fruit, for both significance levels, thus the change in significance levels do not change the conclusion.

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3 years ago
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