What is the equivalent fraction for 3,2,6,4 whole number?

The Average cost of a long distance call in the USA was 31 cents per minute in 1978 and 8 cents in 2008. Find the approximate pe

olya-2409 [2.1K]

Answer:

74%

Step-by-step explanation:

We have been given that the Average cost of a long distance call in the USA was 31 cents per minute in 1978 and 8 cents in 2008.

We will use percent decrease formula to solve our given problem.

$\text{Percent decrease}=\frac{\text{Initial}-\text{Final}}{\text{Initial}}\times 100\%$

$\text{Percent decrease}=\frac{31-8}{31}\times 100\%$

$\text{Percent decrease}=\frac{23}{31}\times 100\%$

$\text{Percent decrease}=0.741935483\times 100\%$

$\text{Percent decrease}=74.1935483\%$

$\text{Percent decrease}\approx 74\%$

Therefore, there is an approximate 74% decrease in cost per minute for the long distance call from 1978 to 2008.

4 0

3 years ago

Solve the equation for the indicated variable.<br><br> Show work, please!!

Tomtit [17]

$A=\dfrac{bh}{2}\\
2A=bh\\
b=\dfrac{2A}{h}$

3 0

3 years ago

Tiffany has eaten 1/4 of her candy bar.William has eaten 2/8 of his candy bar.Who has eaten the greater amount? Explain your ans

zalisa [80]

Neither one has eaten a greater amount, because 1/4 and 2/8 are equal to each other. You know this because when you divide the numerator and the denominator of 2/8 by 2, you get 1/4, which is the exact same thing. The only way the amounts would vary is if the size of one candy bar is greater than the other.

8 0

3 years ago

Read 2 more answers

If I buy a pair of $45 dollar jeans and I get a 15% percent discount but I have to pay a 6% sales a tax how much are the jeans g

irina [24]

$45 x .15 = $6.75 discount
$45 - $6.75= $38.25

Sales tax 6%
$38.25 x 0.6 =$ 2.29
$38.25 + $2.29 = 40.54$

Jeans total is 40.54$

6 0

4 years ago

Read 2 more answers

12. A research study was conducted to estimate the number of white perch (a type of fish) in a Midwestern lake. 300 perch were c

Artyom0805 [142]

Answer:

If the scientist’s estimate about the number of fish in the lake is correct, then it is 44% likely to get 20 perch out of 50 with a tag.

Step-by-step explanation:

Let p be the proportion of tagged white perch in the Midwestern lake.

Scientist's claim is that p= $\frac{300}{1000} =0.30$

Let's test this hypothesis as:

$H_{0}:$ p=0.30

$H_{a}:$ p≠0.30

P-value of the test statistic will give the likelihood of getting 20 perch out of 50 with a tag if the scientist's estimate ( $H_{0}$ ) is true.

Test statistic can be calculated using the equation

$z=\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }$ where

p(s) is the sample proportion of white perch ( $\frac{20}{50} =0.25$ )

p is the proportion assumed under null hypothesis. (0.30)

N is the sample size (50)

Then $z=\frac{0.25-0.30}{\sqrt{\frac{0.30*0.70}{50} } }$ ≈ -0.77

Two tailed p-value of the test statistic is ≈ 0.44

Thus if the scientist’s estimate about the number of fish in the lake is correct, (p=0.30) then it is 44% likely to get 20 perch out of 50 with a tag.

4 0

4 years ago