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Nataly [62]
3 years ago
12

Personal Finance

Mathematics
1 answer:
ch4aika [34]3 years ago
6 0

Answer:

Fresh air

Step-by-step explanation:

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I just need a starter to help me on this. Thanks:)
zhenek [66]
I could help. So mark days as the X value and hours as the Y value
so for the days(x) start with 1 and put in the hours(y) 24 the for the second column in the days(x) box put 2 and in the hours(y) put 48 then continue to at 12 to the hours(y) and 1 to the days(x)
5 0
4 years ago
Mia and Jose shared a pie . Mia are 1/3 and Jose ate 2/6 who ate more
andriy [413]

Answer:

Well both of them ate the same thing.

Step-by-step explanation:

Do 1/3 multiply 2 which is 2/6 and 2/6=2/6 Same thing

5 0
3 years ago
Enter an equation in standard form for the line.<br> Slope is -4, and (-9, -12) is on the line.
AfilCa [17]

Answer:

4x + y +48=0

Step-by-step explanation:

We are here given the slope of the line and one point through which it passes . The slope of the line is -4 and the point is (-9,-12) .

Here we can use the point slope form of the line as ,

\sf\longrightarrow y-y_1 = m(x-x_1)

Substitute the respective values ,

\sf\longrightarrow y - (-12) = -4\{ x -(-9)\}

Simplify LHS and RHS ,

\sf\longrightarrow y + 12 = -4( x +9)

Open the brackets in RHS ,

\sf\longrightarrow y + 12 = -4x -36

Add 36 and 4x to both sides ,

\sf\longrightarrow \underline{\boxed{\orange{\sf  4x + y + 48 =0}}}

8 0
3 years ago
11
Natalka [10]
Your answer is $7.50x=90
3 0
3 years 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
3 years ago
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