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zepelin [54]
3 years ago
13

HELP ME!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Mathematics
2 answers:
Alika [10]3 years ago
7 0
B .................................
anygoal [31]3 years ago
4 0

Answer: B. I am so sorry if i got it wrong please tell me

Step-by-step explanation:

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A bus covers a distance of 90km at a uniform speed. Had the speed been 15km/hour it would have taken 30 minutes less for the jou
Brrunno [24]

Answer:

Step-by-step explanation

Let's say d = distance, r = speed and t = time

We know the distance formula is :

distance = speed x time

d = r multiplied by t

d = rt

If the speed is 15km/hr, the time would be 30 minutes shorter.

Since d = rt,

90 = 15t

t = 6,

This means the time taken for the bus to complete the journey at 15km/hr would be 6hrs.

Originally though, the journey takes 30 minutes longer.

So we can say t = 6.5 hrs , and we know d = 90

Now we use the distance formula:

d = rt, since we want to find the original speed, we want to find r.

Speed = distance/time

r = d/t

r = 90/6.5

r = 180/13

Therefore the original speed is 180/13 km/hr

4 0
3 years ago
Determine whether the equation is always sometimes or never true 2(5x+4)=10x+6
Usimov [2.4K]
This equation is NEVER true. Let's solve the equations algebraically by applying the Distributive Property of Multiplication and transposing the terms across the equality sign. The solution is as follows:

<span>2(5x+4)=10x+6
10x + 8 = 10x + 6
10x - 10x = 6 - 8
0 = -2

Therefore, it is not true.</span>
3 0
3 years ago
Angles X and Y form a linear pair. If the measure of angle X = 3x + 127 and the measure of angle Y = 5x - 11, find the measure o
stiv31 [10]
The answer is C)53 hope this helped !
5 0
3 years ago
Adding 1/2 to 1/4 equals
Alexxx [7]
1/2 + 1/4= 3/4

1/2 + 1/2 = 2/4

2/4+ 1/4= 3/4

6 0
3 years ago
Working for a car company, you have been assigned to find the average miles per gallon (mpg) for acertain model of car. you take
Orlov [11]

Answer:

a) The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b) Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c) 37 cars would have to be sampled.

Step-by-step explanation:

Question a:

We have the sample standard deviation, and thus, the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 15 - 1 = 14

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 2.1448

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 2.1448\frac{6.2}{\sqrt{15}} = 3.4

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 26.7 - 3.4 = 23.3 mpg.

The upper end of the interval is the sample mean added to M. So it is 26.7 + 3.4 = 30.1 mpg.

The 95% confidence interval for the mean mpg, for the certain model of car is (23.3, 30.1). This means that we are 95% sure that the true mean mpg of the model of the car is between 23.3 mpg and 30.1 mpg.

b. What would happen to the interval if you increased the confidence level from 95% to 99%? Explain

Increasing the confidence level, the value of T would increase, thus increasing the margin of error and making the interval wider.

c. The lead engineer is not happy with the interval you constructed and would like to keep the width of the whole interval to be less than 4 mpg wide. How many cars would you have to sample to create the interval the engineer is requesting?

Width is twice the margin of error, so a margin of error of 2 would be need. To solve this, we have to consider the population standard deviation as \sigma = 6.2, and then use the z-distribution.

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Z-table as such z has a p-value of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

Now, find the margin of error M as such

M = z\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

How many cars would you have to sample to create the interval the engineer is requesting?

This is n for which M = 2. So

M = z\frac{\sigma}{\sqrt{n}}

2 = 1.96\frac{6.2}{\sqrt{n}}

2\sqrt{n} = 1.96*6.2

\sqrt{n} = \frac{1.96*6.2}{2}

(\sqrt{n})^2 = (\frac{1.96*6.2}{2})^2

n = 36.9

Rounding up:

37 cars would have to be sampled.

8 0
3 years ago
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