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3 years ago

Which of the following statements describe the graphs of the solution set flr this inequality select two that apply

Mathematics

1 answer:

Helga [31]3 years ago

7 0

Answer:

1st . The shaded region of the graph ....

2nd. The graph line is solid.

4th. The ordered pair (2 ; 5) is part of the solution set.

Step-by-step explanation:

. Because y is greater or equal than the line, everything above the line is part of the solution.

. Since y is greater or equal, the line is solid. If y was only greater the line would have been dotted.

. The pair (2 ; 5) represent the x = 2 w/ y = 5, and reading the graph you can see the at this point the lines passes through it.

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A hand-held digital music player was marked down by 1/4 of the original price. If the sales price is $200. What is the original

Xelga [282]

Answer:A boo

Step-by-step explanation: trust me :) ✨

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2 years ago

How to find the gradient of line segment joining each pair of points?

yarga [219]

Answer:

1/3

Step-by-step explanation:

A(xA, yA) = A (3,1)

B(xB,yB)=B(5,5)

the gradient=(yB-yA)/ (xB-xA)=(5-1)/(5-3)=4/2=2

C(4,7),D(10,9)

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6 0

3 years ago

A company surveyed 2400 men where 1248 of the men identified themselves as the primary grocery shopper in their household. a) E

polet [3.4K]

Answer:

a) With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

b) The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

c) $\alpha =1-0.98=0.02$

Step-by-step explanation:

If np' and n(1-p') are higher than 5, a confidence interval for the proportion is calculated as:

$p'-z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }\leq p\leq p'+z_{\alpha/2}\sqrt{\frac{p'(1-p')}{n} }$

Where p' is the proportion of the sample, n is the size of the sample, p is the proportion of the population and $z_{\alpha/2}$ is the z-value that let a probability of $\alpha/2$ on the right tail.

Then, a 98% confidence interval for the percentage of all males who identify themselves as the primary grocery shopper can be calculated replacing p' by 0.52, n by 2400, $\alpha$ by 0.02 and $z_{\alpha/2}$ by 2.33

Where p' and $\alpha$ are calculated as:

$p' = \frac{1248}{2400}=0.52\\\alpha =1-0.98=0.02$

So, replacing the values we get:

$0.52-2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\leq p\leq 0.52+2.33\sqrt{\frac{0.52(1-0.52)}{2400} }\\0.52-0.0238\leq p\leq 0.52+0.0238\\0.4962\leq p\leq 0.5438$

With a confidence level of 98%, the percentage of all males who identify themselves as the primary grocery shopper are between 0.4962 and 0.5438.

The lower limit of the confidence interval is higher that 0.43, so if he conduct a hypothesis test, he will find that the data shows evidence to said that the fraction is higher than 43%.

Finally, the level of significance is the probability to reject the null hypothesis given that the null hypothesis is true. It is also the complement of the level of confidence. So, if we create a 98% confidence interval, the level of confidence $1-\alpha$ is equal to 98%