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2 years ago

Solve for x. 7 = 3 9 x A) 15 B) 18 C) 21 D) 24

Mathematics

2 answers:

Nataly [62]2 years ago

4 0

Answer:

it is D =)

Step-by-step explanation:

cause i am right believe me i got it wrong because i believed the other guy

ziro4ka [17]2 years ago

3 0

Answer:

I think the answer is D

Step-by-step explanation:

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X - Y =4 - 2x+y=18 Linnear system

Dmitry_Shevchenko [17]

there is no answer to this so i can't help you.?

8 0

2 years ago

Factor 9(x + 4)^2 + (x + 4) – 10 completely. HELP!!!

AleksAgata [21]

Answer:

Step-by-step explanation:

7 0

2 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

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

2 years ago

2. Tessa used the computer every day. Her mother said Tessa could only use the

Vlada [557]

Answer:

3/4 of an hour

Step-by-step explanation:

To find the leftover time, subtract the total allotted time by the time used.

First make sure they have common denominators.

1 1/2 = 3/2 = 6/4

6/4 - 3/4 = 3/4

Tessa has 3/4 of an hour left.

7 0

2 years ago

Cortez's workout plan involves a total of 65 minutes of running and swimming. he burns 11 calories per minute when running and 9

Aleks [24]

The equation to model this would be 65 = 11r + 9s.

In this equation, we know that the total number he works is 65. We also know he burns 11 every time he runs for an hour. Therefore we can multiply 11 by r. We also know he burns 9 for every hour he spends swimming. Therefore we can multiply 9 by s.

When we put all three of these things together, we get 65 = 11r + 9s.

6 0

2 years ago