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Anarel [89]
2 years ago
12

Holly had $5,000 in her bank account. She withdrew $800 to buy a new bike. What is the percent decrease in the balance of her ac

count?
The percent decrease is 1.6%.
The percent decrease is 16%.
The percent decrease is 84%.
The percent decrease is 120%.
Mathematics
2 answers:
ratelena [41]2 years ago
8 0
16% is your Answer 
$5000 *.16=$800
OLga [1]2 years ago
7 0
The percent decrease is 16%. 

Hint: USE THE PERCENT DECREASE FORMULA!
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2x+y=25. 3y=2x-13 solve by elimination
Mila [183]
What  are you trying to find X or Y ?

3 0
1 year ago
Suppose you are told that a distribution is said to be approximately normal because outliers have skewed the data. List the poss
andrew-mc [135]

Answer:

Step-by-step explanation:

Outline are values which are entirely different from those remaining values in a data set. These extreme values can skew an approximately normal distribution by skewing the distribution in the direction of the outliers and this makes it difficult for the data set to be analyzed.

Its effect is such that the mean becomes extremely sensitive to extreme outliers making it possible that the mean is this not a representative of the population and this theoretically affects the standard deviation.

8 0
1 year ago
hugo is saving for a new baseball glove. he saves $10 the first week, and $6 each week for the next 6 weeks. the expression 10 +
BigorU [14]
So if Hugo has saved $10 the first week and $6 for the next 6 weeks the answer would be $46.

you can either add up 6 six times or you can multiple 6 by 6.
6+6+6+6+6+6=36 or 6×6=$36

then you add the $10 from the first week & the $36 from the $6 you saved for 6 weeks to get your final answer of $46.
$36+$10=$46
5 0
1 year ago
Bob's bill for dinner at surefire steak house was $45. In addition he piad 5°\° of the bill in tax and he left a tip for 15°\° o
Anuta_ua [19.1K]

Answer:

$54

Step-by-step explanation:

First

45 times .05= 2.25

Second

45 plus 2.25

Third

45 times .15= 6.75

Fourth

45+2.25+6.75

5 0
1 year ago
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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