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nadezda [96]
3 years ago
6

If the interest rate on a savings account is 0.018%, approximately how much money do you need to keep in this account for 1 year

to earn enough interest to cover a single $9.99 Below-Minimum-Balance Fee?
$55,555
$55
$555
$5,555
Mathematics
1 answer:
krok68 [10]3 years ago
5 0
$55,555 would be the answer
You might be interested in
Someone help please! Determine the intercept of the line that passes through the following points ​
storchak [24]

Answer:

x: (-5.0)

y: (0, -17.5)

Step-by-step explanation:

The x-intercept of the line is when y=0. In the table is the point (-5,0). This is the x-intercept.

To find the y-intercept, find when x=0. Write an equation for the table in y=mx + b. Find the slope between two points first.

m = \frac{y_2-y_1}{x_2-x_1} = \frac{7-0}{-7--5} =\frac{7}{-7+5}=-\frac{7}{2} = -3.5

The slope is -3.5. So the equation is

y - 7 = -3.5(x--7)

y - 7 = -3.5 (x+7)

y - 7 = -3.5x - 24.5

y = 3.5x - 17.5

Since it is in y=mx+b, b= -17.5 and this is the y-intercept.

5 0
3 years ago
Find the area and thank
Lina20 [59]

Answer:

add it

Step-by-step explanation:

3 0
3 years ago
Read 2 more answers
1/5 ( 15 + 10k ) = please help and make sure your answer is accurate.
astra-53 [7]
The answer is 3 + 2k

This is because multiplying by 1/5 is like dividing each term by 5

10k + 15 = 5 • (2k + 3)

8 0
2 years ago
Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobil
o-na [289]

The question is incomplete! Complete question along with answer and step by step explanation is provided below.

Question:

Most individuals are aware of the fact that the average annual repair cost for an automobile depends on the age of the automobile. A researcher is interested in finding out whether the variance of the annual repair costs also increases with the age of the automobile. A sample of 26 automobiles 4 years old showed a sample standard deviation for annual repair costs of $120 and a sample of 23 automobiles 2 years old showed a sample standard deviation for annual repair costs of $100. Let 4 year old automobiles be represented by population 1.

State the null and alternative versions of the research hypothesis that the variance in annual repair costs is larger for the older automobiles.

At a 0.01 level of significance, what is your conclusion? What is the p-value?

Answer:

Null hypotheses = H₀ = σ₁² ≤ σ₂²

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic = 1.44

p-value = 0.1954

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

Step-by-step explanation:

Let σ₁² denotes the variance of 4 years old automobiles

Let σ₂² denotes the variance of 2 years old automobiles

State the null and alternative hypotheses:

The null hypothesis assumes that the variance in annual repair costs is smaller for older automobiles.

Null hypotheses = H₀ = σ₁² ≤ σ₂²

The alternate hypothesis assumes that the variance in annual repair costs is larger for older automobiles.

Alternative hypotheses = Ha = σ₁² > σ₂²

Test statistic:

The test statistic is given by

Test statistic = σ₁²/σ₂²

Test statistic = 120²/100²

Test statistic = 1.44

p-value:

The degree of freedom corresponding to 4 years old automobiles is given by

df₁ = n - 1  

df₁ = 26 - 1  

df₁ = 25

The degree of freedom corresponding to 2 years old automobiles is given by

df₂ = n - 1  

df₂ = 23 - 1  

df₂ = 22

Using Excel to find out the p-value,  

p-value = FDIST(F-value, df₁, df₂)

p-value = FDIST(1.44, 25, 22)

p-value = 0.1954

Conclusion:

When the p-value is less than the significance level then we reject the Null hypotheses

p-value < α   (reject H₀)

But for the given case,

p-value > α

0.1954 > 0.01

Since the p-value is greater than the given significance level therefore, we cannot reject the null hypothesis.

We can conclude that there is no sufficient evidence to support the claim that the variance in annual repair costs is larger for older automobiles.

8 0
3 years ago
Thirty students in the fifth grade class listed their hair and eye colors in the table below: Brown hair Blonde hair Total Green
Mars2501 [29]

Answer:

No, the events "brown hair" and "brown eyes" are not  independent.

Step-by-step explanation:

The table that represents the  hair and eye colors of thirty students of the fifth grade are given in table as:

                         Brown hair          Blonde hair              Total

Green eyes           9                       6                               15

Brown eyes          10                       5                              15

Total                     19                        11                             30

No, the events  "brown hair" and "brown eyes" are not independent.

Since, two events A and B are said to be independent if:

P(A∩B)=P(A)×P(B)

where P denotes the probability of an event.

Here we have:

A= students having brown hair.

B= students having brown eyes.

A∩B= students having both brown hair and brown eyes.

Now,

P(A)=19/30  (ratio of addition of first column to the total entries)

P(B)=15/30  ( ratio of addition of second row to the total entries)

Also,

P(A∩B)=10/30

Now as:

P(A∩B) ≠ P(A)×P(B)

Hence, the two events are not independent.

8 0
2 years ago
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