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natka813 [3]
2 years ago
6

Shay's net earnings per month are $4000. Shay's total monthly expenses are $3000. The biggest expense is rent at $1500. If Shay

takes on a roommate who pays $500 per month, by what amount has Shay reduced total monthly expenses?
Group of answer choices
1/8 or 13%
1/6 or 17%
1/3 or 33%
1/2 or 50%
Mathematics
2 answers:
Gre4nikov [31]2 years ago
8 0
<h3>Answer: Choice B)  1/6 or 17%</h3>

Explanation:

Ignore the net earnings of $4000 as it's extra unneeded info. The same goes for the $1500 rent.

Shay has monthly expenses of $3000. If the roommate chips in $500, then Shay saves 500/3000 = 5/30 = 1/6 of her money toward expenses.

Using a calculator, 1/6 = 0.17 = 17% approximately

Allushta [10]2 years ago
7 0

Answer:1/6 or 17%

Step-by-step explanation:

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Answer:

3y² / x⁴z⁶

Step-by-step explanation:

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ladessa [460]

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3 years ago
The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt
AVprozaik [17]

Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

                             P.Q. = \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample average height = 175 cm

            \sigma = population standard deviation = 20 cm

            n = sample of men = 40

<em>Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.</em>

So, 95% confidence interval for the true mean, \mu is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                     level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } } < 1.96) = 0.95

P( -1.96 \times }{\frac{\sigma}{\sqrt{n} } } < {\bar X-\mu} < 1.96 \times }{\frac{\sigma}{\sqrt{n} } } ) = 0.95

P( \bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } } < \mu < \bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } } ) = 0.95

<u>95% confidence interval for </u>\mu = [ \bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } } , \bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } } ]

                                            = [ 175-1.96 \times }{\frac{20}{\sqrt{40} } } , 175+1.96 \times }{\frac{20}{\sqrt{40} } } ]

                                            = [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

<em>The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.</em>

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3 years ago
For question 2 not 3 :)
BabaBlast [244]

Answer:

2.5

Step-by-step explanation:

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4 years ago
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