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3 years ago

Help could it be any two integers?

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

5 0

Yes, as long as it's between the two numbers

51) -4 and 1

52) -1 and -3

53) -9 and -10

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Elizabeth is waiting for her flight at the airport. She decides to conduct an experiment. She wants to know how many people at t

Likurg_2 [28]

Answer:

the population is everyone at the airport ,the sample is the 50 people that walked by Elizabeth

7 0

4 years ago

If B is the midpoint of AC, and AC= 8x-20, find BC.

Semmy [17]

Answer:

BC = 4x-10

Step-by-step explanation:

$\because B$ is the midpoint of AC.

$\therefore BC = \frac{1}{2} AC$

$\therefore BC = \frac{1}{2} (8x-20)$

$\therefore BC = \frac{1}{2} \times 2(4x-10)$

$\huge \purple {\boxed {\therefore BC = 4x-10}}$

5 0

3 years ago

Look at the screen shot

Zepler [3.9K]

Answer:

M- MONICA

Step-by-step explanation:

6 0

2 years ago

Which of the following best describes a bisector of an angle

Gnom [1K]

Answer:

An angle only has one bisector. Each point of an angle bisector is equidistant from the sides of the angle. The interior or internal bisector of an angle is the line, half-line, or line segment that divides an angle of less than 180° into two equal angles.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Anas wants to build a one-sample z interval with 85% confidence to estimate what proportion of users will click an advertisement

klasskru [66]

Answer:

See below

Step-by-step explanation:

Check One Sample Z-Interval Conditions

Simple Random Sample? √

np≥10? √

n(1-p)≥10? √

One-Sample Z-Interval Information

Formula --> $CI=\hat{p}\pm z^*\biggr(\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\biggr)$
Sample Proportion --> $\hat{p}=\frac{54}{250}=0.216$
Critical Value --> $z^*=1.4395$ (for a 85% confidence level)
Sample Size --> $n=250$
Margin of Error (MOE) --> $\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

Problem 1

As stated previously, Anas should use the critical value $z^*=1.4395$ to construct the 85% confidence interval

Problem 2

Given our formula for the margin of error (MOE), the value is $MOE=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.216(1-0.216)}{250}}\approx0.026$

Problem 3

The 85% confidence interval would be $CI=\hat{p}\pm z^*\biggr(\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\biggr)=CI=0.216\pm 1.4395(0.026)\appro=\{0.1786,0.2534\}$ , which means that we are 85% confident that the true proportion of people that clicked on the advertisement is between 0.1786 (~45 people) and 0.2534 (~63 people)

Problem 4

Increasing the sample size to $n=500$ is going to decrease the margin of error because it is a closer representation of the population, but, alas, requires more time, energy, and resources to observe.

8 0

2 years ago