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3 years ago

I don't understand the math problem, this is for a first grader

Mathematics

1 answer:

icang [17]3 years ago

5 0

Looks like it wants it in groups maybe?

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What is the value of x?<br> a. 2 /29<br> b. 3 /61<br> c. 2/ 101<br> d. 2/333

stiv31 [10]

Answer:

2√133

Step-by-step explanation:

a²+b²=c²

8²+12²=c²

c²=208

208+18²=532

√532=23

3 0

3 years ago

A scientest is studying a new super bacteria that is threatening to cause an outbreak. On the first day, there were 5 organisms

ElenaW [278]

Answer:

C. 125

Step-by-step explanation:

Given:

Bacteria on day 1 = 5

On day 2, each bacteria got split into 5.

So, number of bacteria on day 2 = $5\times 5 = 25$

Now, again on day 3, each of the 25 bacteria got split into additional 5 bacteria.

So, number of of bacteria on day 3 = $25\times 5 =125$

Therefore, the population of bacteria on day 3 is 125.

8 0

3 years ago

What is the solution for 13.2 = t 4.4? I'm super like spazzed out right now and can't think.

Olegator [25]

Answer:

4.4 t = 13.2

t = 13.2 / 4.4 = 3

t = 3

Step-by-step explanation:

5 0

3 years ago

In the diagram below, triangle ABC is drawn with altitude BD. If AD = 30 and AC = 36, determine the length of AB rounded to the

denis-greek [22]

9514 1404 393

Answer:

32.9

Step-by-step explanation:

The ratio of long side to hypotenuse is the same for all of the triangles in the figure.

AD/AB = AB/AC

AB^2 = AD·AC = 30·36

AB = 6√30 ≈ 32.863

To the nearest tenth, AB ≈ 32.9.

5 0

3 years ago

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population pro

Gnom [1K]

Answer:

A sample of 1068 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

The margin of error is:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$ , so $Z = 1.96$ .

At 95% confidence, how large a sample should be taken to obtain a margin of error of 0.03 for the estimation of a population proportion?

We need a sample of n.

n is found when M = 0.03.

We have no prior estimate of $\pi$ , so we use the worst case scenario, which is $\pi = 0.5$

Then

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.03 = 1.96\sqrt{\frac{0.5*0.5}{n}}$

$0.03\sqrt{n} = 1.96*0.5$

$\sqrt{n} = \frac{1.96*0.5}{0.03}$

$(\sqrt{n})^{2} = (\frac{1.96*0.5}{0.03})^{2}$

$n = 1067.11$

Rounding up

A sample of 1068 is needed.

8 0

3 years ago