Hi anyone tell me the answer

Based on the family the graph below belongs to, which equation could represent the graph? image below.

natka813 [3]

Answer:

y = 1/x+2 + 3

Step-by-step explanation:

x = -2

y = 3

8 0

4 years ago

Assume that the readings on the thermometers are normally distributed with a mean of 0 degrees and standard deviation of 1.00deg

IrinaVladis [17]

Answer:

Step-by-step explanation:

Hello!

The variable of interest is the readings on thermometers. This variable is normally distributed with mean μ= 0 degrees C and standard deviation σ= 1.00 degrees C.

The objective is to find the readings that are in the top 3.3% of the distribution and the lowest 3.3% of the distribution.

Symbolically:

The lower value P(X≤a)=0.033

Top value P(X≥b)=0.033

(see attachment)

Lower value:

The accumulated probability until "a" is 0.03, since the variable has a normal distribution, to reach the value of temperature that has the lowest 3.3%, you have to work under the standard normal distribution.

First we look the Z value corresponding to 0.033 of probability:

Z= -1.838

Now you reverste the standardization using the formula Z= (a-μ)/δ

a= (Z*δ)+μ

a= (-1.838*1)+0

a= -1.838

Top value:

P(X≥b)=0.033

This value has 0.033 of the distribution above it then 1 - 0.033= 0.967

is below it.

You can rewrite the expression as:

P(X≤b)=0.967

Now you have to look the value of Z that corresponds to 0.967 of accumulated probability:

b= (Z*δ)+μ

b= (1.838*1)+0

b= 1.838

The cutoff values that separates rejected thermometers from the others are -1.838 and 1.838 degrees C.

I hope it helps!

5 0

3 years ago

If f(1)=2f(1)=2 and f(n)=f(n-1)^2-n

katrin [286]

Answer:

f(4)=-3

Step-by-step explanation:

f(1)=2

f(n)=f(n-1)^2-n

If n=2

f(2)=f(2-1)^2-2

f(2)=f(1)^2-2

f(2)=2^2-2

f(2)=4-2

f(2)=2

If n=3

f(3)=f(3-1)^2 - 3

f(3)=f(2)^2 - 3

f(3)=2^2-3

f(3)=4-3

f(3)=1

if n=4

f(4)=f(4-1)^2 - 4

f(4)=f(3)^2 - 4

f(4)=1^2 - 4

f(4)=1-4

f(4)=-3

6 0

3 years ago

Each side of a sandbox is 6 feet long. It will cost $3.00 per square foot to replace the sand in the sandbox. What would be the

omeli [17]

Answer:

I don't know

Step-by-step explanation:

Unknown, but you could probably use Googl#

3 0

3 years ago

"Immediately after a ban on using hand-held cell phones while driving was implemented, compliance with the law was measured. A r

sergiy2304 [10]

Answer:

(a) Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1= p_2$

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$

(b) We conclude that there is a statistical difference in these two proportions measured initially and then one year later.

Step-by-step explanation:

We are given that a random sample of 1,250 drivers found that 98.9% were in compliance. A year after the implementation, compliance was again measured to see if compliance was the same (or not) as previously measured.

A different random sample of 1,100 drivers found 96.9% compliance."

Let $p_1$ = proportion of drivers that were in compliance initially

$p_2$ = proportion of drivers that were in compliance one year later

(a) Null Hypothesis, $H_0$ : $p_1-p_2=0$ or $p_1= p_2$ {means that there is not any statistical difference in these two proportions measured initially and then one year later}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq 0$ or $p_1\neq p_2$ {means that there is a statistical difference in these two proportions measured initially and then one year later}

The test statistics that will be used here is Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{ \frac{\hat p_1(1-\hat p_1)}{n_1} + \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of drivers in compliance initially = 98.9%

$\hat p_2$ = sample proportion of drivers in compliance one year later = 96.9%

$n_1$ = sample of drivers initially = 1,250

$n_2$ = sample of drivers one year later = 1,100

(b) So, the test statistics = $\frac{(0.989-0.969)-(0)}{\sqrt{ \frac{0.989(1-0.989)}{1,250} + \frac{0.969(1-0.969)}{1,100}} }$

= 3.33

Now, P-value of the test statistics is given by;

P-value = P(Z > 3.33) = 1 - P(Z $\leq$ 3.33)

= 1 - 0.99957 = 0.00043

Since in the question we are not given with the level of significance so we assume it to be 5%. Now at 5% significance level, the z table gives critical values between -1.96 and 1.96 for two-tailed test.

Since our test statistics does not lies within the range of critical values of z, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that there is a statistical difference in these two proportions measured initially and then one year later.

7 0

3 years ago