All categories

3 years ago

45%, 15%, 40%, 80%, 65%, 30%, 90%, 25% Do any of these percentages match the fraction 2/5?

Mathematics

2 answers:

Slav-nsk [51]3 years ago

8 0

Answer:

Yes, the third option (40%)

Step-by-step explanation:

Given the fraction with numerator 2 and denominator 5:

$\frac{2}{5}$

You can convert it from this fraction form to a decimal number by dividing the numerator 2 by the denominator 5, obtaining the decimal number:

$\frac{2}{5}=0.4$

Finally, to rewrite this decimal number to percentage, you need to multiply it by 100. Then:

$0.4*100=40\%$

Therefore, there is a percentage that matches the fraction $\frac{2}{5}$ . This percentage is 40%, which corresponds to the third option.

stellarik [79]3 years ago

3 0

Answer:

Yes

$\frac{2}{5} \: matches \: 40\%$

Step-by-step explanation :

To change a fraction to percentage, multiply the fraction by 100%

This implies that

$\frac{2}{5} \times 100\%$
We cancel out the common factors to obtain,

$2 \times 20\%$

This simplifies to

$= 40\%$

Hence

$\frac{2}{5} \: matches \: 40\%$

You might be interested in

Tia and her 5 friends are going to the ice skating ring. Each person pays $5 for admission and $5 for food. Write a numerical ex

dedylja [7]

The expression would be (5*6)+(5*6)=x
or you could shorten it and do (10*6)=x

8 0

3 years ago

1 point<br> Captionless Image

algol [13]

Answer:

uhhh ok

Step-by-step explanation:

6 0

3 years ago

The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal arc welds used in LNG ship co

Alona [7]

Answer:

95% Confidence interval for the variance:

$3.6511\leq \sigma^2\leq 34.5972$

95% Confidence interval for the standard deviation:

$1.9108\leq \sigma \leq 5.8819$

Step-by-step explanation:

We have to calculate a 95% confidence interval for the standard deviation σ and the variance σ².

The sample, of size n=8, has a standard deviation of s=2.89 miles.

Then, the variance of the sample is

$s^2=2.89^2=8.3521$

The confidence interval for the variance is:

$\dfrac{ (n - 1) s^2}{ \chi_{\alpha/2}^2} \leq \sigma^2 \leq \dfrac{ (n - 1) s^2}{\chi_{1-\alpha/2}^2}$

The critical values for the Chi-square distribution for a 95% confidence (α=0.05) interval are:

$\chi_{0.025}=1.6899\\\\\chi_{0.975}=16.0128$

Then, the confidence interval can be calculated as:

$\dfrac{ (8 - 1) 8.3521}{ 16.0128} \leq \sigma^2 \leq \dfrac{ (8 - 1) 8.3521}{1.6899}\\\\\\3.6511\leq \sigma^2\leq 34.5972$

If we calculate the square root for each bound we will have the confidence interval for the standard deviation:

$\sqrt{3.6511}\leq \sigma\leq \sqrt{34.5972}\\\\\\1.9108\leq \sigma \leq 5.8819$

6 0

3 years ago

The 1997 value of an object was $9500. In 2012, it was worth $5000. The annual percent of decay has been constant. What is the a

vova2212 [387]

Hello,

Here is your answer:

The proper answer to this question is option D "4.19%".

Your answer is D.

If you need anymore help feel free to ask me!

Hope this helps!

5 0

3 years ago

Read 2 more answers

Cholesterol levels for a group of women aged 30-39 follow an approximately normal distribution with mean 190.14 milligrams per d

Anettt [7]

Answer:

Step-by-step explanation:

Hello!

X: Cholesterol level of a woman aged 30-39. (mg/dl)

This variable has an approximately normal distribution with mean μ= 190.14 mg/dl

1. You need to find the corresponding Z-value that corresponds to the top 9.3% of the distribution, i.e. is the value of the standard normal distribution that has above it 0.093 of the distribution and below it is 0.907, symbolically:

P(Z≥z₀)= 0.093

-*or*-

P(Z≤z₀)= 0.907

Since the Z-table shows accumulative probabilities P(Z<Z₁₋α) I'll work with the second expression:

P(Z≤z₀)= 0.907

Now all you have to do is look for the given probability in the body of the table and reach the margins to obtain the corresponding Z value. The first column gives you the integer and first decimal value and the first row gives you the second decimal value:

z₀= 1.323

Using the Z value from 1., the mean Cholesterol level (μ= 190.14 mg/dl) and the Medical guideline that indicates that 9.3% of the women have levels above 240 mg/dl you can clear the standard deviation of the distribution from the Z-formula:

Z= (X- μ)/δ ~N(0;1)

Z= (X- μ)/δ

Z*δ= X- μ

δ=(X- μ)/Z

δ=(240-190.14)/1.323

δ= 37.687 ≅ 37.7 mg/dl

I hope it helps!

8 0

3 years ago