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baherus [9]
3 years ago
9

Jeremy is going on rides at a carnival he starts the rides a 12:30 by 12:45 he had gone on 1/5 of the carnival rides at the rate

how many hours will it take him to go on all the rides at the carnival
PLS answer i need it fast!
Mathematics
1 answer:
MArishka [77]3 years ago
8 0

Answer:

1.25 hours or 1 hour and 15 minutes

Step-by-step explanation:

12:45 - 12:30 = :15

15 minutes equates to 1/5 of the rides

So, we multiply 15 by 5 to find the total time taken to ride all the rides.

15 x 5 = 75

75/60 = 1.25

It would take him 1.25 hours to ride all the rides at the carnival or 1 hour and 15 minutes.

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0.15(y-0.2)=2-0.5(1-y)<br><br> The answer is -153/35 I just need to know how to get that
butalik [34]

Answer:

y=-\dfrac{153}{35}.

Step-by-step explanation:

The given equation is

0.15(y-0.2)=2-0.5(1-y)

Using distributive property, we get

0.15(y)+0.15(-0.2)=2-0.5(1)-0.5(-y)

0.15y-0.03=2-0.5+0.5y

0.15y-0.03=1.5+0.5y

Isolate variable terms.

0.15y-0.5y=1.5+0.03

-0.35y=1.53

Divide both sides by -0.35.

y=\dfrac{1.53}{-0.35}

y=-\dfrac{1.53}{0.35}\times \dfrac{100}{100}

y=-\dfrac{153}{35}

Therefore, y=-\dfrac{153}{35}.

6 0
3 years ago
When the sample size and the sample proportion remain the same, a 90 percent confidence interval for a population proportion p w
deff fn [24]

Answer:

A 90% confidence interval for <em>p</em> will be <u>narrower </u>than the 99% confidence interval.

Step-by-step explanation:

The formula to compute the (1 - <em>α</em>) % confidence interval for a population proportion is:

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

Here \hat p is the sample proportion.

The margin of error of the confidence interval is:

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

The MOE is dependent on:

  1. Confidence level
  2. Standard deviation
  3. Sample size

The MOE is directly related to the confidence level and standard deviation.

So if any of the two increases then the MOE also increases, thus widening the confidence interval.

And the MOE is inversely related to the sample size.

So if the sample increases the MOE decreases and vice versa.

It is provided that the sample size and the sample proportion are not altered.

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

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

And the critical value of <em>z</em> for 99% confidence level is:

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

So as the confidence level increases the critical value increases.

Thus, a 90% confidence interval for <em>p</em> will be narrower than the 99% confidence interval.

6 0
2 years ago
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soldier1979 [14.2K]

Answer:

b is the answer I choose because it is going in-5

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Step-by-step explanation:

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let's firstly convert the mixed fractions to improper fractions, and then subtract.


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5 0
2 years ago
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