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n200080 [17]
3 years ago
7

A friendly contest involves randomly drawing a marble from a bag that contains 16 blue marbles, 12 red

Mathematics
1 answer:
matrenka [14]3 years ago
7 0

Answer: I have a huge feeling it would be Nathan because he has the most blue marble. Yk?

You might be interested in
17. Simplify (3 × 22) ÷ 6 + [28 – (4)2] = ?
Effectus [21]

Answer:

31

Step-by-step explanation:

(3*22=66) / 6 + [28n - (4)2=20]

66 /6 = 11+ [20]

11+20= 31

7 0
3 years ago
Find the solution to the inequality<br><br> 23(6x−3)&gt;6?
laiz [17]

Answer:

The answer is x>25/46

Step-by-step explanation:

Distribute 23 through the parentheses

138x-69>6

Move constant to the the right and change the sign. 138x>6+69.

Add the numbers like this: 138x>75

Divide both sides and you'll will get x>25/46 which is your answer. Let me know if this helps

8 0
2 years ago
A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the
EastWind [94]

Answer:

a) \hat p=\frac{471}{1024}=0.460

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

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".  

The population proportion have the following distribution

p \sim N(p,\sqrt{\frac{p(1-p)}{n}})

Data given and notation  

n=1024 represent the random sample taken    

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

\hat p=\frac{471}{1024}=0.460 estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering    

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

The standard error is given by:

SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156

And the margin of error is given by:

ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305

Part b

If we replace the values obtained we got:

0.460-1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.429

0.460+1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.491

The 99% confidence interval would be given by (0.429;0.491)

8 0
2 years ago
Find the direction angle of 3i+1j.<br><br> Please help me. Thank you
Alekssandra [29.7K]

Answer:

Correct answer:  ∝ = 18.43°

Step-by-step explanation:

Given coordinates represent one vector in x-y plane, let be v.

v = 3i +1j = xi + yj => x = 3 and y = 1

The angle that this vector builds with the positive direction to the x axis will be found using the tangent.

∝ = tan⁻¹ y/x = tan⁻¹ 1/3 = tan⁻¹ 0.333 = 18.43°

∝ = 18.43°

God is with you!!!

4 0
2 years ago
PLS HELP! <br> find the volume.
Andre45 [30]
Don’t do that it’s a scam and a virus
7 0
2 years ago
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