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3 years ago

The fraction four-fifths is equivalent to what percent?

Mathematics

2 answers:

iren [92.7K]3 years ago

7 0

<h2>Hello!</h2>

The answer is:

The fraction is equivalent to 80%.

To solve the problem, we need to remember that if we want to convert from numbers to percentual values, we need to multiply the given number by 100.

So, if we have the fraction four-fifths which is:

$\frac{4}{5}=0.8$

If we need to convert it to percent, we need to multiply it by 100:

$0.8*100=80(percent)$

Hence, we have that the fraction is equivalent to 80%.

Have a nice day!

erica [24]3 years ago

7 0

Answer:

80%

Step-by-step explanation:

We know 4/5 is the same as 4 divided by 5 then using long division for 4 divided by 5 gives us 0.8 converting our number to a percent

0.8 x 100 = 80%

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Which of the following is not part of polyhedron.

Rom4ik [11]

Answer:

Step-by-step explanation:

Oblique. The other three are used to describe a polyhedron. They are related by the following formula

V + F - E = 2.

Where

V = number of vertexes
F = the number of faces
E = edges

For example consider a common cube, like a sugar cube.

It has 6 faces (f)

It has 8 vertexes (V)

It has 12 edges. E

6 + 8 - 12 = 14 - 12 = 2

8 0

3 years ago

<br><br> f(1)=−6<br> f(2)=−4<br> f(n)=f(n−2)+f(n−1)<br> <br> f(n)=?

andriy [413]

The nth term of the sequence is 2n - 8

<h3>Equation of a function</h3>

The nth term of an arithmetic progression is expressed as;

Tn = a + (n - 1)d

where

a is the first term

d is the common difference

n is the number of terms

Given the following parameters

a = f(1)=−6

f(2) = −4

Determine the common difference

d = f(2) - f(1)

d = -4 - (-6)
d = -4 + 6

d = 2

Determine the nth term of the sequence

Tn = -6 + (n -1)(2)

Tn = -6+2n-2
Tn = 2n - 8

Hence the nth term of the sequence is 2n - 8

Learn more on nth term of an AP here: brainly.com/question/19296260

#SPJ1

5 0

1 year ago

Read 2 more answers

Emily's bill for dinner at a restaurant was $61. She left an 18% tip. What was the amount of the tip?

UkoKoshka [18]

You cross multiply. So 61 over 18 = x over 100. You multiply 61 and get the answer of 1600. Now with what you have left, you divide that with 18 and you should get your answer of <span>338.888889. Make sure to round it to whatever it tells you to round to!</span>

7 0

3 years ago

Jovana is inviting 9 friends to a party each friend wants 7 cookies and each box has 12 cookies how many boxes should Jovana get

gulaghasi [49]

Answer:

3boxes

Step-by-step explanation:

1st we should know the number of cookies

7*5=35

Divide the total of cookies to cookies per box 35/12=2 11/12

2 boxes and 11 cookies remaning (they need an extra box)= 3 boxes

8 0

3 years ago

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. What perce

LenKa [72]

Answer:

14.28% of individual adult females have weights between 75 kg and 83 kg.

92.82% of the sample means are between 75 kg and 83 kg.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

Assume that weights of adult females are normally distributed with a mean of 79 kg and a standard deviation of 22 kg. This means that $\mu = 79, \sigma = 22$ .

What percentage of individual adult females have weights between 75 kg and 83 kg?

This percentage is the pvalue of Z when $X = 83$ subtracted by the pvalue of Z when $X = 75$ . So:

X = 83

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{83 - 79}{22}$

$Z = 0.18$

$Z = 0.18$ has a pvalue of 0.5714.

X = 75

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{75- 79}{22}$

$Z = -0.18$

$Z = -0.18$ has a pvalue of 0.4286.

This means that 0.5714-0.4286 = 0.1428 = 14.28% of individual adult females have weights between 75 kg and 83 kg.

If samples of 100 adult females are randomly selected and the mean weight is computed for each sample, what percentage of the sample means are between 75 kg and 83 kg?

Now we use the Central Limit THeorem, when $n = 100$ . So $s = \frac{22}{\sqrt{100}} = 2.2.$