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3 years ago

What is the length of a car if the length on the scale model car is 17 cm and the scale is 2 cm : 50 cm?

Mathematics

2 answers:

alexira [117]3 years ago

5 0

Answer:

Answers 425 is you multiply 50 by 8 you’d get 400 then add 25 the number 8 is the amount of times you can use the scale so it would be 16cm them you have one more cm left so you’d just cut the scale in half

Step-by-step explanation:

Zarrin [17]3 years ago

3 0

Read more about area at:

brainly.com/question/24487155

#SPJ1

3 0

1 year ago

Please help me, will mark brainliest for correct answers.....

Alexandra [31]

Answer:

1. B) 5.7

2. A) 12

3. A) 11.4

4. A) 5.7

5. A) 16.2

6. A) 11.2

7. No, they do not form a right triangle

8. Yes, they do form a right triangle

Step-by-step explanation:

Extra tip: The hypotenuse has to be less than both sides added together, but cannot be more than either of the sides alone.

16² + b² = 17²

256 + b² = 289

256 - 256 + b² = 289 - 256

b² = 33

√b² = √33

b = 5.74 or 5.7

16² + b² = 20²

256 + b² = 400

256 - 256 + b² = 400 - 256

b² = 144

√b² = √144

b = 12

7² + 9² = c²

49 + 81 = c²

130 = c²

√130 = √c²

11.40 or 11.4 = c

7² + b² = 9²

49 + b² = 81

49 - 49 + b² = 81 - 49

b² = 32

√b² = √32

b = 5.65 or 5.7

a² + 5² = 17²

a² + 25 = 289

a² + 25 - 25 = 289 - 25

a² = 264

√a² = √264

a = 16.24 or 16.2

10² + b² = 15²

100 + b² = 225

100 - 100 + b² = 225 - 100

b² = 125

√b² = √125

b = 11.18 or 11.2

15² + 8² = 16²

225 + 64 = 256

289 ≠ 256

5² + 12² = 13²

25 + 144 = 169

169 = 169

8 0

3 years ago

I need help due today at 6

Fofino [41]

Answer:

Step-by-step explanation:

3 0

3 years ago

All of the following are examples of essay types that colleges typically aska student to write except O A. essays about you. B.

Nataliya [291]

Answer: C. essays about work experience

Step-by-step explanation: I would chose C because when you apply for a job you are asked to write and an essay on your work experience. But in college, you are no applying for a job.

7 0

2 years ago

The local oil changing business is very busy on Saturday mornings and is considering expanding. A national study of similar busi

eimsori [14]

Answer:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts and data given

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".

$\bar X=4.2$ represent the sample mean

$s=1.4$ represent the sample standard deviation

n=16 represent the sample selected

$\alpha=0.05$ significance level

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is different from 3.6, the system of hypothesis would be:

Null hypothesis: $\mu = 3.6$

Alternative hypothesis: $\mu \neq 3.6$

If we analyze the size for the sample is < 30 and we know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{4.2-3.6}{\frac{1.4}{\sqrt{16}}}=1.714$

Critical values

On this case since we have a bilateral test we need to critical values. We need to use the t distribution with $df=n-1=16-1=15$ degrees of freedom. The value for $\alpha=0.05$ and $\alpha/2=0.025$ so we need to find on the t distribution with 15 degrees of freedom two values that accumulates 0.025 of the ara on each tail. We can use the following excel codes:

"=T.INV(0.025,15)" "=T.INV(1-0.025,15)"

And we got $t_{crit}=\pm 2.131$

So the decision on this case would be:

Reject the null hypothesis if the observed "t" value is less than -2.131 or higher than 2.131

Rejection Zone: $t_{calculated}$ or $t_{calculated}>2.131$

Conclusion

In our case since our calculated value is not on the rejection zone we don't have enough evidence to reject the null hypothesis at 5% of significance.

3 0

3 years ago