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2 years ago

9

Help me please ............................

1 answer:

denpristay [2]2 years ago

4 0

Answer:

3/2

Step-by-step explanation:

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(10x-11) + (3x-2) + (3x+1) = 180

Sauron [17]

Answer:

right across my mom is the best I can get in a lot and I don't think so

Step-by-step explanation:

carbon fiber and fiber is not a problem in the morning and it doesn't work for me because I am sorry to get the best results in the process but it will get worse when I have a good idea to make a difference and 87inches will be there for me and I have no problems in my body step and the symptoms of my illness will have 7 or so severe symptoms that may 6PM up with my symptoms of depression and anxiety and depression in my heart as a child in a very much and very much

8 0

2 years ago

Read 2 more answers

10. In a survey of 212 people at the local track and field championship, 72% favored the home team

igomit [66]

Answer:

a. The margin of error for the survey is of 0.0308 = 3.08%.

b. The 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning is (65.96%, 78.04%).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error of the survey is:

$M = \sqrt{\frac{\pi(1-\pi)}{n}}$

The confidence interval can be written as:

$\pi \pm zM$

In a survey of 212 people at the local track and field championship, 72% favored the home team winning.

This means that $n = 212, \pi = 0.72$

a. Find the margin of error for the survey.

$M = \sqrt{\frac{0.72*0.28}{212}} = 0.0308$

The margin of error for the survey is of 0.0308 = 3.08%.

b. Give the 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning.

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

Lower bound:

$\pi - zM = 0.72 - 1.96*0.0308 = 0.6596$

Upper bound:

$\pi + zM = 0.72 + 1.96*0.0308 = 0.7804$

As percent:

0.6596*100% = 65.96%

0.7804*100% = 78.04%.

The 95% confidence interval that is likely to contain the exact percent of all people who favor the home team winning is (65.96%, 78.04%).

7 0

3 years ago

What is the solution to the equation below?

MakcuM [25]

Answer:

X=4

Step-by-step explanation:

The solution is in the file

5 0

2 years ago

Read 2 more answers

Please helpppp ty giving brain liest trying to get more:)

olya-2409 [2.1K]

The answer I believe would be 3/4

Remember to count from point to point using rise/run

I hope this helped :D

8 0

2 years ago

Read 2 more answers

Given that the mass of an average linebacker at Ursinus College is 250 lbs and the radius of a pea is 0.50 cm, calculate the num

Brilliant_brown [7]

Answer:

1.544*10⁹ Linebackers would be required in order to obtain the same density as an alpha particle

Step-by-step explanation:

Assuming that the pea is spherical ( with radius R= 0.5 cm= 0.005 m), then its volume is

V= 4/3π*R³ = 4/3π*R³ = 4/3*π*(0.005 m)³ = 5.236*10⁻⁷ m³

the mass in that volume would be m= N*L (L= mass of linebackers=250Lbs= 113.398 Kg)

The density of an alpha particle is ρa= 3.345*10¹⁷ kg/m³ and the density in the pea ρ will be

ρ= m/V

since both should be equal ρ=ρa , then

ρa= m/V =N*L/V → N =ρa*V/L

replacing values

N =ρa*V/L = 3.345*10¹⁷ kg/m³ * 5.236*10⁻⁷ m³ /113.398 Kg = 1.544*10⁹ Linebackers

N=1.544*10⁹ Linebackers

7 0

3 years ago