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2 years ago

Please help fast!!!!

Mathematics

1 answer:

emmasim [6.3K]2 years ago

7 0

I think it’s 3 not sure!

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What is 2/9 divided by 3?

Kryger [21]

((2÷9))÷3
(0.222)÷3
=0.074

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3 years ago

Find m<1 then m<2. State the theroms or prostulates that justify your answers

ki77a [65]

Answer:

fffgfdfgee

Step-by-step explanation:

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3 years ago

Read 2 more answers

Convert: 42 centigrams 6 milligrams = 1 milligrams

Naddik [55]

Answer:

centigram is 'bigger.'

Well, if you look on the metric scale it is apparent that 1 mg = 0.1 (or 1/10) cg suggesting that a milligram is the smaller value.

6 0

3 years ago

kate takes 48 minutes to get to the park. if she gets to the park at 2:32 what time did she leave for the park

olya-2409 [2.1K]

Answer:

1:44

Step-by-step explanation:

You can subtract 32 minutes from 2:32 to get 2:00.

Since you only subtracted 32, subtract 16 (16+32=48) from 60, since 60 minutes is one hour, to get 1:44

8 0

2 years ago

A truck loaded with 8000 electronic circuit boards has just pulled into a firm’s receiving dock. The supplier claims that no mor

Juliette [100K]

Answer:

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

Step-by-step explanation:

Let X = number of boards that fall outside the most rigid level of industry performance specifications.

In a random sample of 300 boards the number of defective boards was 12.

Compute the sample proportion of defective boards 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%).

3 0

2 years ago