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3 years ago

Help I’ll give 50 points 12) 4x- y=8 5x+y=1

Mathematics

1 answer:

Luba_88 [7]3 years ago

4 0

12) 4x- y=8

5x+y=1

Correct.

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Who ever help I will give you the brainliest answer

Gala2k [10]

Answer:

just do it Like Nikey

Step-by-step explanation:

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4 0

2 years ago

60 is 30% of what number?

Phantasy [73]

Percent means parts out of 100
30%=30/100=3/10

'of' means multiply

60 is 30% of what translates to
60=3/10 times what
multiply both sides by 10/3
600/3=what
200=what
the number is 200

7 0

3 years ago

Which set of numbers could represent the lengths of the sides of a right triangle?

DiKsa [7]

Answer:

The first set: 8, 15, and 17.

Step-by-step explanation:

By the pythagorean theorem, a triangle is a right triangle if and only if

$\text{longest side}^2 = \text{first shorter side}^2 + \text{second shorter side}^2$ .

In this case,

$\text{longest side}^2 = 17^2 = 289$ .

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 8^2 + 15^2\\ &=64 + 225 = 289 \end{aligned}$ .

In other words, indeed $\text{hypotenuse}^2 = \text{first leg}^2 + \text{second leg}^2$ . Hence, 8, 15, 17 does form a right triangle.

Similarly, check the other pairs. Keep in mind that the square of the longest side should be equal to the sum of the square of the two

Factor out the common factor $2$ to simplify the calculations.

$\text{longest side}^2 = 20^2 = 400$

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 10^2 + 15^2\\ &=100 + 225 = 325 \end{aligned}$ .

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

$\text{longest side}^2 = (2\times 11)^2 = 2^2 \times 121$ .

$\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= (2 \times 6)^2 + (2 \times 9)^2\\ &=2^2 \times(36 + 81) = 2^2 \times 117 \end{aligned}$ .

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

$\text{longest side}^2 = 11^2 = 121$ .

<h3> $\begin{aligned}&\text{first shortest side}^2 + \text{second shortest side}^2 \\ &= 7^2 + 9^2\\ &=49+ 81 = 130 \end{aligned}$ .</h3>

$\text{longest side}^2 \ne \text{first shorter side}^2 + \text{second shorter side}^2$ .

Hence, by the pythagorean theorem, these three sides don't form a right triangle.

6 0

3 years 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

<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

A health food store has launched a new pack of vitamins for children. The store records the number of packs ordered by a random

Kisachek [45]

Answer:

The Answer is the second one

Step-by-step explanation:

3 0

2 years ago