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fiasKO [112]
2 years ago
13

Bo Karah can run a race in 4 hours 33 minutes if he runs at 7 m/s. How long will it take in hours and minutes if

Mathematics
1 answer:
lidiya [134]2 years ago
8 0

Answer:

7 hours and 9 minutes

Step-by-step explanation:

4 hours and 33 minutes is 273 minutes

7 m/s —> 273 minutes

11 m/s —> M

Criss cross to get M

11x273= 7M

3003=7M

M=429 minutes

==> 429 minutes is 7 hours and 9 minutes

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Tickets to a show cost $6.50 for adults and $4.75 for students. A family is purchasing 2 adult tickets and 3 student tickets.
GuDViN [60]

Answer:

1) The estimate would be about $24

2) The exact cost would be $27.25

3) They should receive $2.75

Hope this helps :)

7 0
2 years ago
A ferris wheel can accommodate 54 people in 2/5 hours.
barxatty [35]
54/2=27
27•5=135
135 people can ride the Ferris wheel per hour
135•3=405
405 people can ride the Ferris wheel in 3 hours
4 0
2 years ago
Nana can pack a bag of groceries in 6/8 of a minute, at that rate how many can she pack in 15 minutes?
rjkz [21]
So 1=6/8 minute
1=3/4 minute
multiply both sides by 4
4=3 minutes

3 times what=15
divide by 3
what=5

4=3 minutes
mulitpliy both sides by 5
20=15 minutes

answer is 20 bags

5 0
3 years ago
What is the average rate of change from x = -1 x = 2 equal to 5?
kherson [118]
Yes. And thanks.for free branana
7 0
2 years ago
Information about the proportion of a sample that agrees with a certain statement is given below. Use StatKey or other technolog
lara31 [8.8K]

Answer:

SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048

If we replace the values obtained we got:

0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257

0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443

The 95% confidence interval would be given by (0.257;0.443)

Step-by-step explanation:

Notation and definitions

X=35 number of people that agree.

n=100 random sample taken

\hat p=\frac{35}{100}=0.35 estimated proportion of people that agree

p true population proportion of peopl that agree

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}})

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value would be given by:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the mean is given by the following formula:  

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

The standard error is given by:

SE=\sqrt{\frac{\hat p (1-\hat p)}{n}}=\sqrt{\frac{0.35 (1-0.35)}{100}}=0.048

If we replace the values obtained we got:

0.35 - 1.96\sqrt{\frac{0.35(1-0.35)}{100}}=0.257

0.42 + 1.96\sqrt{\frac{0.42(1-0.42)}{150}}=0.443

The 95% confidence interval would be given by (0.257;0.443)

6 0
3 years ago
Read 2 more answers
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