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4 years ago

11

Solve the problem. Make sure your answer is in simplest form and improper fractions are expressed as mixed numbers. 1/5 x 6/7?i

need help plz!:)

2 answers:

allsm [11]4 years ago

6 0

$\it \dfrac{1}{5} \cdot \dfrac{6}{7} = \dfrac{1\cdot6}{5\cdot7} = \dfrac{6}{35}$

Valentin [98]4 years ago

5 0

The answer is 6/35. Just multiply straight across

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The results of Accounting Principals’ latest Workonomix survey indicate the average American worker spends $1092 on coffee annua

Nataly_w [17]

Answer:

a)

Group 18-34 years old

$\bar x = 1041.625 \\ s^2=485301 \\ s=696.635$

Group 35-44 years old

$\bar x = 1359.5 \\ s^2=178548 \\ s=422.549$

Group 45 and older

$\bar x = 1414.375 \\ s^2=18292.27 \\ s=135.248$

b)

According to the sample there is 9.04% probability that a person between 18 and 34 consume less than the average, 47.74% probability that a person between 35 and 44 consume more than the average and 50% probability that a person older than 45 consume more than the average.

Step-by-step explanation:

a)

The <em>mean</em> for each sample is

$\bar x=\frac{\sum_{k=1}^{10}x_k}{10}$

where the $x_k$ are the data corresponding to each group

The <em>variance</em> is

$s^2=\frac{\sum_{k=1}^{10}(\bar x-x_k)^2}{9}$

and the <em>standard deviation </em>is s, the square root of the variance.

<u>Group 18-34 years old </u>

$\bar x = 1041.625 \\ s^2=485301 \\ s=696.635$

<u>Group 35-44 years old </u>

$\bar x = 1359.5 \\ s^2=178548 \\ s=422.549$

<u>Group 45 and older </u>

$\bar x = 1414.375 \\ s^2=18292.27 \\ s=135.248$

b)

Let's compare these averages against the general media established of $1,092 by using the corresponding z-scores

$z=\frac{\bar x-\mu}{s/\sqrt{n}}$

where

<em> $\bar x$ = mean of the sample </em>

<em> $\mu$ = established average </em>

<em>s = standard deviation of the sample </em>

<em>n = size of the sample </em>

<u>z-score of Group 18-34 years old </u>

$z=\frac{1041.625-1092}{696.635/\sqrt{10}}=-0.2286$

The area under the normal curve N(0;1) between -0.2286 and 0 is 0.0904. So according to the sample there is 9.04% probability that a person between 18 and 34 consume less than the average.

<u>z-score of Group 35-44 years old </u>

$z=\frac{1359-1092}{422.5491/\sqrt{10}}=2.0019$

The area under the normal curve N(0;1) between 0 and 2.0019 is 0.4774. So according to the sample there is 47.74% probability that a person between 35 and 44 consume more than the average.

<u>z-score of Group 45 and older </u>

$z=\frac{1414.375-1092}{135.2489/\sqrt{10}}=7.5375$

The area under the normal curve N(0;1) between 0 and 7.5375 is 0.5. So according to the sample there is 50% probability that a person older than 45 consume more than the average.

3 0

4 years ago

Grace is comparing cell phone plans. A prepaid phone plan costs $0.20 per minute and has no monthly fee. A contracted phone plan

Sophie [7]

Answer:

B

Step-by-step explanation:

The contracted phone plan will have the same steepness and a higher y-intercept.

4 0

3 years ago

BRAINLIEST ASAP!!!!!!!!!!!!!!!!!!!!!!!!! (please check my answer)

tatiyna

F(x) = 3x - 16x - x + 4
f(x) = - 14x + 4

I can not see your picture

6 0

3 years ago

Help.....................

Mazyrski [523]

I think abc is equal to QPR

8 0

3 years ago

What situation should be used to rewrite 16(x^3+1)^2 -22(x^3+1)-3=0 as a quadratic equation

Oliga [24]

$16(x^3+1)^2-22(x^3+1)-3=0$

Use substitution: $x^3+1=t$

$16t^2-22t-3=0\\\\16t^2-24t+2t-3=0\\\\8t(2t-3)+1(2t-3)=0\\\\(2t-3)(8t+1)=0\iff2t-3=0\ \vee\ 8t+1=0$

$2t-3=0\ \ \ |+3\\\\2t=3\ \ \ |:2\\\\t=\dfrac{3}{2}\\..............................\\8t+1=0\ \ \ \ |-1\\\\8t=-1\ \ \ \ |:8\\\\t=-\dfrac{1}{8}$

we're going back to substitution:

$x^3+1=\dfrac{3}{2}\ \vee\ x^3+1=-\dfrac{1}{8}\ \ \ \ |subtract\ 1\ from\ both\ sides\ of\ the\ equations\\\\x^3=\dfrac{1}{2}\ \vee\ x^3=-\dfrac{9}{8}\\\\x=\sqrt[3]{\dfrac{1}{2}}\ \vee\ x=\sqrt[3]{-\dfrac{9}{8}}$

$x=\dfrac{1}{\sqrt[3]2}\ \vee\ x=-\dfrac{\sqrt[3]9}{2}\\\\\boxed{x=\dfrac{\sqrt[3]4}{2}\ \vee\ x=-\dfrac{\sqrt[3]9}{2}}$

4 0

3 years ago