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3 years ago

14 mm 29 mm Calculate the surface area.

Mathematics

1 answer:

tino4ka555 [31]3 years ago

4 0

Answer:

14x29=406 2hich would be the area

Step-by-step explanation:

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Caroline is trying to earn money to pay for her field trip by doing odd jobs around the house. She needs to raise more than $35

Alenkasestr [34]

The inequality that represents how much money Caroline needs to earn will be m>35.

In this case, Caroline needs an amount that's greater than $35, therefore, the amount must be greater than $35

From the options given, the correct inequality will be m>35. This means m is greater than $35.

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8 0

2 years ago

.Shaun estimated that the attendance at a college

Serggg [28]

Answer:

38.72%

Step-by-step explanation:

Estimated attendance = 1,000

The actual value of attendance = 1,632

We need to find the percent error of Shaun's estimate. Using the formula as follows :

$\%=\dfrac{|\text{actual value-estimated value}|}{\text{actual value}}\times 100$

Put all the values,

$\%=\dfrac{1632-1000}{1632}\times 100\\\\=38.72\%$

So, the percent error of Shaun's estimate is 38.72%.

8 0

3 years ago

What is the mean of the following set of data? {6,9,3, 8, 9,3,4}

Setler79 [48]

Answer:

Step-by-step explanation:

Add all the numbers together and divide by the amount of numbers

6+9+3+8+9+3+4= 42

42/7 = 6

7 0

3 years ago

Read 2 more answers

Est 100: 337 55,745 SS t 1,000:

Andre45 [30]

Answer:

Is this a question??

Step-by-step explanation:

7 0

3 years ago

A software developer wants to know how many new computer games people buy each year. Assume a previous study found the standard

Marrrta [24]

Answer:

The minimum sample size required to ensure that the estimate has an error of at most 0.14 at the 95% level of confidence is n=567.

Step-by-step explanation:

We have to calculate the minimum sample size n needed to have a margin of error below 0.14.

The critical value of z for a 95% confidence interval is z=1.96.

To do that, we use the margin of error formula in function of n:

$MOE=\dfrac{z\cdot \sigma}{\sqrt{n}}\\\\\\n=\left(\dfrac{z\cdot \sigma}{MOE}\right)^2=\left(\dfrac{1.96\cdot 1.7}{0.14}\right)^2=(23.8)^2=566.42\approx 567$

The minimum sample size to have this margin of error is n = 567.

3 0

3 years ago