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2 years ago

At any point in time, there could be bicycles, tricycles, and

Mathematics

1 answer:

Aleksandr-060686 [28]2 years ago

6 0

Answer:

There may be 1 or 3 tricycles in the parking lot.

Step-by-step explanation:

Since at any point in time, there could be bicycles, tricycles, and cars in the school parking lot, and today, there are 53 wheels in total, if there are 15 bicycles, tricycles, and cars in total, to determine how many tricycles could be in the parking lot, the following calculation must be performed:

13 x 4 + 1 x 3 + 1 x 2 = 57

11 x 4 + 1 x 3 + 3 x 2 = 53

10 x 4 + 3 x 3 + 2 x 2 = 53

8 x 4 + 5 x 3 + 2 x 2 = 51

10 x 2 + 1 x 3 + 4 x 4 = 39

9 x 3 + 1 x 2 + 5 x 4 = 49

Therefore, there may be 1 or 3 tricycles in the parking lot.

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Step-by-step explanation:

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2 years ago

Simplify (8m2+m+7m2)-(3m-4m3+2m2)

kondaur [170]

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3 years ago

Make the following changes to the number 5.376: Make the 3 worth 10 times as much. Makethe5worth_1 asmuch.

cluponka [151]

Answer:

Step-by-step explanation:

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3 years ago

Trinh is serving 112 people cake , 5/8ths are getting chocolate how many people will get carrot ?

Novay_Z [31]

Hey!

If 112 people are getting cake and 5/8 are getting chocolate, that means the remaining will get chocolate.

8/8 - 5/8 = 3/8

3/8 are getting chocolate. You have to find 3/8 of 112.

3/8 × 112 = 336/8 = 42

42 people are getting carrot cake

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5 0

2 years ago

A team averaging 110 points is likely to do very well during the regular season. The coach of your team has hypothesized that yo

aev [14]

Answer:

1. M=108

2. μ=110

3. In the explanation.

4. Test statistic t = -1.05

5. P-value = 0.1597

Step-by-step explanation:

The question is incomplete: to solve this problem, we need the sample information: size, mean and standard deviation.

We will assume a sample size of 10 matches, a sample mean of 108 points and a sample standard deviation of 6 points.

1. The mean points is the sample points and has a value of 108 points.

2. The null hypothesis is H0: μ=110, meaning that the mean score is not significantly less from 110 points.

3. This is a hypothesis test for the population mean.

The claim is that the mean score is significantly less than 110.

Then, the null and alternative hypothesis are:

$H_0: \mu=110\\\\H_a:\mu< 110$

The significance level is 0.05.

The sample has a size n=10.

The sample mean is M=108.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=6.

The estimated standard error of the mean is computed using the formula:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{6}{\sqrt{10}}=1.9$

4. Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{108-110}{1.9}=\dfrac{-2}{1.9}=-1.05$

The degrees of freedom for this sample size are:

$df=n-1=10-1=9$

5. This test is a left-tailed test, with 9 degrees of freedom and t=-1.05, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.1597) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the mean score is significantly less than 110.

7 0

3 years ago