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1 year ago

What is the area of this shape? 12 cm A 15 cm 5 cm 2 cm

Mathematics

1 answer:

MrRa [10]1 year ago

3 0

Answer: 84.

Step-by-step explanation: If you add the width 5cm and 2cm, it gives you 7cm, then length x width so 12x7= 84(cm).

If I am wrong, I am truly sorry, I haven't done area in a long time.

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A family drove 165 miles in 3 hours. At the same rate, how many miles could they travel in 8 hours

Oksi-84 [34.3K]

First you have to divide 165 by 3, which is 55. and then 55 multiplied by 8 is 440. so in 8 hours, they could travel 440 miles.

6 0

4 years ago

Read 2 more answers

Please show work! Please and thank you!

My name is Ann [436]

Question 5(A)
Answer:51
Explanation
Step 1:evaluate the power
Step2:add the numbers
Step 3:calculate the square root
Step4:reduce the fraction with 4
Step5:calculate the sum or the difference and ur final answer is 51

Question 5(B)
Answer:9/2
Explanation:
Step 1:calculate the difference
Step 2:evaluate the power
Step 3:subtract the numbers
Step 4:calculate the sum
And ur final answer is 9/2

Question 6:(A)
Answer:x=36/11
Explanation:
Step 1:distribute 4 through the parentheses
Step 2:move variable to the left hand side and change its sign
Step 3:move constant to the right hand side and Change it’s sign
Step 4:collect like items
Step 5:add the numbers
Step 6:divide both sides of the equation by 11
And ur final answer is x=36/11

Question 6(B)
Answer:x=192
Explanation:
Step 1:multiply both sides of the equation by 3
Step 2:move the constant to the right hand side and change its sign
Step 3:add the numbers
And ur final answer is x=192

Question 7
Answer B($60)
Explanation:the shape is rectangular and it is 12 feet long and the other is 5 feet wide it’s like finding the area of it so u basically need to multiply 12 by 5 and ur answer is 60

Here is the answers and the explanations step by step I hope it helps
Have a great day and stay safe

6 0

3 years ago

Find each sum (-4y+3) + (11y-5)

Airida [17]

Answer:

7y−2

Step-by-step explanation:

8 0

3 years ago

An article in The Engineer (Redesign for Suspect Wiring," June 1990) reported the results of an investigation into wiring errors

GarryVolchara [31]

Answer:

a) The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b) A sample of 408 is required.

c) A sample of 20465 is required.

Step-by-step explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

Of 1600 randomly selected aircraft, eight were found to have wiring errors that could display incorrect information to the flight crew.

This means that $n = 1600, \pi = \frac{8}{1600} = 0.005$

99% confidence level

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $Z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 - 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0005$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.005 + 2.575\sqrt{\frac{0.005*0.995}{1600}} = 0.0095$

The 99% confidence interval on the proportion of aircraft that have such wiring errors is (0.0005, 0.0095).

b. Suppose we use the information in this example to provide a preliminary estimate of p. How large a sample would be required to produce an estimate of p that we are 99% confident differs from the true value by at most 0.009?

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A sample of n is required, and n is found for M = 0.009. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.009 = 2.575\sqrt{\frac{0.005*0.995}{n}}$

$0.009\sqrt{n} = 2.575\sqrt{0.005*0.995}$

$\sqrt{n} = \frac{2.575\sqrt{0.005*0.995}}{0.009}$

$(\sqrt{n})^2 = (\frac{2.575\sqrt{0.005*0.995}}{0.009})^2$

$n = 407.3$

Rounding up:

A sample of 408 is required.

c. Suppose we did not have a preliminary estimate of p. How large a sample would be required if we wanted to be at least 99% confident that the sample proportion differs from the true proportion by at most 0.009 regardless of the true value of p?

Since we have no estimate, we use $\pi = 0.5$