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ololo11 [35]
3 years ago
7

After paying a $10 admission fee, you walk into a carnival. A sign reads "$3 per ride". When your parent picks you up, they aske

d you how much you spent, and you told them $52. WRITE AN EQUATION that will help you answer the question of how many rides you went on. Do NOT hit the spacebar. Do NOT use capital letters. All fractions must be simplified (no mixed numbers - keep those as improper fractions) PLEASE HELP
Mathematics
1 answer:
alina1380 [7]3 years ago
8 0

9514 1404 393

Answer:

  10+3x=52

Step-by-step explanation:

If you went on x rides, your cost for the rides is 3x. That added to the $10 admission fee gives you the total you spent:

  3x+10=52

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A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the
lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

Step-by-step explanation:

(a)

Let <em>X</em> = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

<em>X</em> = 610

<em>n</em> = 1000

Confidence level = 90%

The (1 - <em>α</em>)% confidence interval for population proportion is:

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

The sample proportion is:

\hat p=\frac{X}{n}=\frac{610}{1000}=0.61

The critical value of <em>z</em> for a 90% confidence level is:

z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

<u>Interpretation</u>:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let <em>X</em> = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

<em>X</em> = 440

<em>n</em> = 1000

Confidence level = 95%

The (1 - <em>α</em>)% confidence interval for population proportion is:

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

The sample proportion is:

\hat p=\frac{X}{n}=\frac{440}{1000}=0.44

The critical value of <em>z</em> for a 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend  is (0.42, 0.46).

<u>Interpretation</u>:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

  1. The confidence level
  2. Sample size
  3. Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of <em>z</em> in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

4 0
3 years ago
You run one lap around a mile track every 8 minuets. Your friend runs around the same track every 10 minutes. You both start at
GalinKa [24]
So 8x5=40 and 10×4=40 so you and your friend meet at the 40 th min
6 0
3 years ago
I need help with 19 20 21
lord [1]
19) 5c+c=-6+2
      6c=-4
      c=-4/6
      c=-2/3

20) -20-8=-2e
       -28=-2e
       28/2=e
      14=e

21)p-2w=2l
     (p-2w)/2=l
5 0
3 years ago
10) The temperature recorded at 12 noon was 15°C above zero. If it decreases at the rateof 3°C per hour until midnight, at what
aalyn [17]

Answer:

at 8 pm the temperature will be -9 degrees. and at midnight the temperature will be negitive twenty one(-21) degrees

Step-by-step explanation:

the start time is 12 pm and finish is 12 am and every hour your going to subtract 3 from the given degree so 15-3= 12 , 12-3=9 ect.

6 0
3 years ago
How to solve -5 + 13y = -7
zubka84 [21]

Answer:

y = -2/13

Step-by-step explanation:

-5 + 13y = -7

Add 5 to each side

-5+5 + 13y = -7+5

13y = -2

Divide by 13

13y/13 = -2/13

y = -2/13

5 0
3 years ago
Read 2 more answers
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