Which of the following best approximates the line of the best fit

Solve thank you I will give brainlist to first

Gre4nikov [31]

Answer:

Step-by-step explanation:

no because each y value doesn’t increase uniformly

4 0

3 years ago

Read 2 more answers

Abigail was skateboarding home when the wheel axle of her skateboard broke. She had already traveled two thirds of the way home

OverLord2011 [107]

Answer:

I think that her taking the skateboard is 4 times as fast as her walking.

Step-by-step explanation:

I got this because she traveled twice as long and twice as fast on the skateboard, so if you multiply it together, there u have it.

4 0

3 years ago

Suppose that the number of a certain type of computer that can be sold when its price is P (in dollars) is given by a linear fun

Scilla [17]

Answer:

a) N(P) = -6P + 16000

b) slope = -6 computers per dollar

That means the number of computer sold reduce by 6 per dollar increase in price.

c) ∆N = -660 computers

Step-by-step explanation:

Since N(P) is a linear function

N(P) = mP + C

Where m is the slope and C is the intercept.

Case 1

N(1000) = 10000

10000 = 1000m + C ....1

Case 2

N(1700) = 5800

5800 = 1700m + C ....2

Subtracting equation 1 from 2

700m = 5800 - 10000

m = -4200/700

m = -6

Substituting m = -6 into eqn 1

10000 = (-6)1000 + C

C = 10000+ 6000 = 16000

N(P) = -6P + 16000

b) slope = -6 computers per dollar

That means the number of computer sold reduce by 6 per dollar increase in price.

Slope is the change in number of computer sold per unit Change in price.

c) since slope m = -6 computers per dollar

∆P = 110 dollars

∆N = m × ∆P

Substituting the values,

∆N = -6 computers/dollar × 110 dollars

∆N = -660 computers.

The number of computer sold reduce by 660 when the price increase by 110 dollars

7 0

3 years ago

Which of the following is the solution to 9|x +4|>54 ?

horsena [70]

Answer:

x>2 or x<-10

Step-by-step explanation:

9|x +4|>54

Divide each side by 9

9/9|x +4|>54/9

|x +4|>6

There are two solutions, one positive and one negative

x+4 >6 or x+4 < -6

Subtract 4 from each side

x+4-4 >6-4 or x+4-4 < -6-4

x>2 or x<-10

4 0

3 years ago

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

vampirchik [111]

Answer:

a) Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b) The 95% confidence interval would be given by (910.05;959.95)

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) $z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750$

Since is a bilateral test the p value is given by:

$p_v =2*P(Z>2.750)=0.0059$

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=935$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=180$ represent the population standard deviation

n=200 represent the sample size

The confidence interval for the mean is given by the following formula:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=3278.222$

The sample deviation calculated $s=97.054$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$