All categories

3 years ago

Whoever gets it right can have brainliest, but answer ASAP!

Mathematics

2 answers:

notka56 [123]3 years ago

8 0

Answer:

512

Step-by-step explanation:

512

dedylja [7]3 years ago

4 0

Answer:

$beinggreat78~here~to~help!$

512

The small 3 in front of 8 is the term, cubed. Cubed means to multiply the number behind it by itself 3 times. That lucky number is 8! Multiply 8 by itself 3 times.

$8^3~or~ 8 * 8 * 8 = 512$

The problem is solved! Your answer is 512

You might be interested in

What might weigh 20 kg?

KatRina [158]

The heavy suitcase.

A small car would still weigh quite a lot, a tablet would weigh a lot less than that, and a watermelon is around 2kg (depends).

7 0

3 years ago

Read 2 more answers

I rly need help on this ty

REY [17]

Answer:

Oh I didn’t know that lol

Step-by-step explanation:

Lol

4 0

3 years ago

Please answer the following:

ratelena [41]

Answer:

$( \frac{7}{10} - \frac{5}{8} )$ ÷ $\frac{1}{16}$

(check your other question for my explanation)

5 0

3 years ago

Find the exact solution of x. x2 = 16

vichka [17]

The answer is x=16/2
x=8

5 0

4 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

3 years ago