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4 years ago

What is the perimeter of this rectangle?

Mathematics

1 answer:

saveliy_v [14]4 years ago

5 0

Answer:3.2 units

Step-by-step explanation:Since the perimeters of the rectangles are equal:

n + n + 0.6 = n +0.1 + 2n

2n + 0.6 = 3n + 0.1

0.5 = n

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Four times a number, minus 9, is equal to three times the number, plus 6

marshall27 [118]

4 x p - 9 = 3 x p + 6
Hope it helped

7 0

3 years ago

Read 2 more answers

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

<em>Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.</em>

<u>So, 80% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

<u>80% confidence interval for </u> $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

<u>98% confidence interval for </u> $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago

Bonita spent $8.94 on groceries. She bought a gallon of milk for $4.29 and 3pounds of sliced turkey. How much does 1 pound of sl

Vesna [10]

<h2>Information</h2>

$8.94 spent in total on groceries

Gallon of milk = $4.29

3 pounds of sliced turkey bought = ?

<h2>Explanation</h2>

Subtract the total amount spent on groceries with what items you have prices for (in this case the gallon of milk):

$8.94 (total spent) - $4.29 (gallon of milk) = $4.65 (for 3 pounds of sliced turkey)

Next, divide the price for 3 pounds of sliced turkey with however many pounds Bonita bought (in this case, 3 pounds):

$4.65 (total for 3 pounds of turkey) ÷ 3 (pounds of turkey) = $1.55 (per pound of turkey)

<h2>Solution</h2>

Each pound of turkey costs $1.55

5 0

3 years ago

SGA is doing a fundraiser to sell candles for project graduation. Small candles cost $4.00 and large candles cost $6.00. They ha

mafiozo [28]

Given:

Cost of small candles = $4.00

Cost of large candles = $6.00

They have at most 50 candles, and want to make at least $100.

To find:

The system of linear inequalities represents the situation.

Solution:

Let x be the number of small candles and y be the number of large candles.

At most 50 candles means total candles must be less than or equal to 50.

$x+y\leq 50$

Want to make at least $100. It means, the total sales must be greater than or equal to $100.

$4x+6y\geq 100$

Therefore, the inequalities in the system of linear inequalities are $x+y\leq 50$ and $4x+6y\geq 100$ .

3 0

3 years ago

Can someone help me plz

Ivan

Wouldent the answer be A because all percents are out of 100 so the 1.9 would be 9 over 100

6 0

3 years ago

What is the perimeter of this rectangle?​

What is the perimeter of this rectangle?