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xenn [34]
2 years ago
12

​Megan hughes deposits $4300 in an account that pays simple interest. when she withdraws her money 7 months later, she receives

$4,450.50. what rate of interest did the account pay? round to the nearest whole percent.
Mathematics
1 answer:
Juliette [100K]2 years ago
3 0

The interest is 3.5 percent or 3 percents of the sum she deposited.

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What are the solution(s) to the quadratic equation 40 − x2 = 0?
Anni [7]
X = 20
First, you move the 40 to the other side.
-2x=-40
Then, you divided -40 by -2.
x= -40/-2
x= 20
x is 20 because the negatives cancel out making it positive.
4 0
2 years ago
Read 2 more answers
Type the correct answer in the box. Round your answer to two decimal places.
Goshia [24]

Answer:

39.3701 inches in a meter

4 0
1 year ago
a school district requires all graduating seniors to take a mathematics test. This year, the rest scores were approximately norm
Klio2033 [76]
A score of 85 would be 1 standard deviation from the mean, 74.  Using the 68-95-99.7 rule, we know that 68% of normally distributed data falls within 1 standard deviation of the mean.  This means that 100%-68% = 32% of the data is either higher or lower.  32/2 = 16% of the data will be higher than 1 standard deviation from the mean and 16% of the data will be lower than 1 standard deviation from the mean.  This means that 16% of the graduating seniors should have a score above 85%.
3 0
2 years ago
While on a trip, you notice that the video screen on the airplane, in addition to showing movies and news, records your altitude
mixas84 [53]

Answer:

It descends 3 km.

Step-by-step explanation:

The rate of change is the slope. The formula for slope is

m=\frac{y_2-y_1}{x_2-x_1}

Using the first two points, (0, 12) and (2, 10), we have

m = (10-12)/(2-0) = -2/2 = -1

This means the plane descends 1 km per minute.  This means in 3 minutes' time, the plane descends 3 km.

3 0
1 year ago
Read 2 more answers
Fowle Marketing Research, Inc., bases charges to a client on the assumption that telephone surveys can be completed in a mean ti
ser-zykov [4K]

Answer:

a)  Null hypothesis:  \mathbf{H_0 : \mu \leq 15}

Alternative hypothesis: \mathbf{H_1 = \mu > 15}

b) the test statistics is : 2.15

c) The p-value is 0.0158

d) NO, there is evidence that the mean time of telephone survey is less than 15 and premium rate is not justified.

Step-by-step explanation:

The data in the  Microsoft Excel are:

17;11;12;23;20;23;15;

16;23;22;18;23;25;14;

12;12;20;18;12;19;11;

11;20;21;11;18;14;13;

13;19; 16;10;22;18;23.

a) Formulate the null and alternative hypotheses for this application.

From the question, Fowle Marketing Research Inc. is taking base charge from a client on the given assumption that if the mean time of telephone survey is 15 minutes or less.

The null and alternative hypotheses are therefore as follows:

Null hypothesis:  \mathbf{H_0 : \mu \leq 15}

The null hypothesis states that there is evidence that the mean time of telephone survey is less than 15 and premium rate is not justified.

Alternative hypothesis: \mathbf{H_1 = \mu > 15}

The alternative hypothesis states that there is evidence that the mean time of telephone survey exceeds 15 and premium rate is justified.

b) Compute the value of the test statistic.

Given that:

\mu = 15

\sigma = 3.6

n = 35

The sample mean \bar x = \dfrac{ \sum x}{n}  is;

\bar x = \dfrac{ 17+11+12 ... 22+18+23}{35}

\bar x = 17

Thus:

z = \dfrac{ \bar  x - \mu }{\dfrac{\sigma}{\sqrt{n}}}

z = \dfrac{ 17 - 15}{\dfrac{3.6}{\sqrt{15}}}

z = \dfrac{ 2}{0.9295}}

\mathbf{z =2.15}

Thus; the test statistics is : 2.15

c) What is the p-value?

p-value = P(Z > 2.15)

p-value = 1 - P(Z ≤ 2.15)

From the standard normal table, the value of P(Z ≤ 2.15) is 0.9842

p-value = 1 - 0.9842

p-value = 0.0158

The p-value is 0.0158

d)  At a = .01, what is your conclusion?

According to the rejection rule, if p-value is less than 0.01 then reject null hypothesis at ∝ = 0.01

Thus; p-value =  0.0158 >  ∝ = 0.01

By the rejection rule, accept the null hypothesis.

Therefore, there is evidence that the mean time of telephone survey is less than 15 and premium rate is not justified.

4 0
2 years ago
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