Can y’all answer the question for me please

Mathematics

1 answer:

Pie3 years ago

8 0

It’s the second one, cuz that’s the correct formula and equation

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In a survey about a new immigration law in Georgia, "46% of the 132 Georgia farmers, agricultural processors, and farm service b

Triss [41]

Answer:

The 132 who responded is the SAMPLE.

Step-by-step explanation:

In a survey about a new immigration law in Georgia, "46% of the 132 Georgia farmers, agricultural processors, and farm service businesses who responded said they were experiencing some degree of labor shortage"

So, here the 132 who responded are samples.

Sample is defined as a small part of a surveyed population, that shows what the whole population must be like.

8 0

3 years ago

Kelly has 4 times as many songs on her music player as Lou. Tiffany has 6 times as many songs on her music player as Lou. Altoge

slava [35]

Let the number of songs owned by Lou be = x.

Kelly has 4 times as many songs as Lou = 4x

Tiffany has 6 times as many songs as Lou = 6x

They all have total of 682 songs

= x + 4x + 6x = 682
   11x = 682
   x = 682 / 11 .    Use your calculator.
   x =62.

How many songs Kelly has = 4x = 4*62 = 248 songs.
Kelly = 248 songs.

6 0

3 years ago

Read 2 more answers

Which ratios are less than 8 to 10? Check all that apply.

kenny6666 [7]

3:5 and 50/100
Are both less than 8:10

16/20 is the same as 8:10 and 13/15 is more than

6 0

3 years ago

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A supervisor records the repair cost for 25 randomly selected dryers. A sample mean of $93.36 and standard deviation of $19.95 a

Feliz [49]

Answer:

The 98% confidence interval for the mean repair cost for the dryers is (83.4161, 103.3039).

Step-by-step explanation:

We have a small sample size n = 25, $\bar{x} = 93.36$ and s = 19.95. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ where $t_{\alpha/2}$ is the $\alpha/2$ th quantile of the t distribution with n - 1 = 25 - 1 = 24 degrees of freedom. As we want the 98% confidence interval, we have that $\alpha = 0.02$ and the confidence interval is $93.36\pm t_{0.01}(\frac{19.95}{\sqrt{25}})$ where $t_{0.01}$ is the 1st quantile of the t distribution with 24 df, i.e., $t_{0.01} = -2.4922$ . Then, we have $93.36\pm (-2.4922)(\frac{19.95}{\sqrt{25}})$ and the 98% confidence interval is given by (83.4161, 103.3039).