Which of the following is a repeating decimal when converted to decimal form?

5 x + 4y= 16 X+y = 3.5 Find out what each letter is equal to X= Y=

klemol [59]

X=4
Y=-0.5

Do you need an explanation?

5 0

3 years ago

Natalie uses 68 photos to make an online photo album. She can fit 5 photos on each page. She fills all of the pages in her album

Sati [7]

68-2(\5)=#of pages in album

6 0

4 years ago

Joan needs to estimate the size of her bedroom so that she can buy enough paint to cover the walls. Two of the walls measure 2.8

labwork [276]

The total area of the room is 37.6376 and the no. of cans required to paint the wall is 3 cans.

The measurement of two of the walls is 2.86 metres and 3.16 metre

Area of the two walls = 2(length x breadth)

Area = 2(2.86 x 3.16) = 18.0752 m²

The measurement of the other two walls is 2.86 metres and 3.42 metres

Area of the two walls = 2(length × breadth)

Area = 2(2.86 × 3.42) = 19.5624 m²

Total area = 18.0752 + 19.5624 = 37.6376 m²

If one can of paint can cover 15 m², the no. of cans required to paint the bedroom will be

No. of cans = Total area/Area covered by one can of paint

No. of cans = 37.6376/15 = 2.5091 = 3 cans (approx.)

3 0

1 year ago

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

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

So, 80% confidence interval for the population mean, $\mu$ is ;

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

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

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

= [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%.

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

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

= [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%.

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

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

= [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

What is the solution of <img src="https://tex.z-dn.net/?f=%5Cfrac%7Bx%2B2%7D%7B4%7D%20%3D%20%5Cfrac%7Bx%2B8%7D%7B3%7D" id="TexFo

Lina20 [59]

Answer:

x = -26

Step-by-step explanation:

Multiply both sides by the lowest common multiple of 4 and 3, which is 12.

3(x+2) = 4(x+8)

3x + 6 = 4x + 32 --- distribute the 4 and 3

6 = x + 32 --- subtract 3x from both sides

-26 = x --- subtract 32 from both sides

x = -26

6 0

3 years ago