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2 years ago

Solve this please how do you select a image on brainly >~

Mathematics

1 answer:

Paladinen [302]2 years ago

3 0

Answer:

CLICK THE ADD ATTACHMENT

Step-by-step explanation:

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Pls help asapppppp!!

denis23 [38]

Answer:

254

Step-by-step explanation:

first find the diameter of the big circle by adding the 2 founded numbers

which is 18

then the Radius = 9

to find area,

$\pi{(r) }^{2}$

which is

$81\pi \: or \: 254$

8 0

2 years ago

Read 2 more answers

The figure shows how the yard lines on a football field are numbered. The goal lines are labeled G. A referee was standing on a

Drupady [299]

YOU WILL DO YOU GOOD LUCK

Step-by-step explanation:

7 0

2 years ago

Which of these methods will produce the most representive sample of the population of cars made by this company

maks197457 [2]

Answer:

The manufacturer tests every 100 car in the assembly line

4 0

2 years ago

Find the new amount given the original amount and the percent of change. $5; 40% increase The new amount is $

bekas [8.4K]

Answer:

Step-by-step explanation:

5$ -40%=3

7 0

2 years ago

Read 2 more answers

In a simple random sample of 300 boards from this shipment, 12 fall outside these specifications. Calculate the lower confidence

Lyrx [107]

Answer:

The 95% confidence interval for the percentage of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

Step-by-step explanation:

In a random sample of 300 boards the number of boards that fall outside the specification is 12.

Compute the sample proportion of boards that fall outside the specification in this sample as follows:

$\hat p =\frac{12}{300}=0.04$

The (1 - α)% confidence interval for population proportion p is:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The critical value of z for 95% confidence level is,

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

Compute the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification as follows:

$CI=\hat p\pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.04\pm1.96\sqrt{\frac{0.04(1-0.04)}{300}}\\=0.04\pm0.022\\=(0.018, 0.062)\\\approx(1.8\%, 6.2\%)$

Thus, the 95% confidence interval for the proportion of all boards in this shipment that fall outside the specification is (1.8%, 6.2%).

6 0

2 years ago