All categories

3 years ago

How many zeros are in the standard form of six hundred thousand, twenty? Explain

Mathematics

1 answer:

creativ13 [48]3 years ago

3 0

Standard form is the literal standard form you write numbers in, so it would be 600,020. Hope this helps!

You might be interested in

Solve the equation <br><br> 3x-5< 8x-1/2<br><br> x+7≤ 4x+1

vladimir1956 [14]

This is all I got. Hope this helps:

6 0

2 years ago

Read 2 more answers

Billy watched the clown car drive up to the circus and saw 14 clowns get out of one little car.The clowns made up 56% of perform

Ganezh [65]

Answer:

there was 25 performers

Hope this helps ❤️

8 0

3 years ago

Read 2 more answers

Use the distance formula to find The distance between the points (-1 ,3) and (2,6) write you answear as a square root, do not gi

olga nikolaevna [1]

Answer:

d = $3\sqrt{2}$

Step-by-step explanation:

The distance formula is d = $\sqrt{(x_{2}-x_{1})^{2}+(y_{2} -y_{1} )^{2} }$

Plug in the points.

d = $\sqrt{(2-(-1))^{2}+(6 -3 )^{2} }$

Solve

d = $\sqrt{18}$

Simplify by separating 18 into 9x2

d = $\sqrt{9(2)}$

Take the square root of 9 to finish simplifying

d = $3\sqrt{2}$

6 0

3 years ago

When fishing off the shores of Florida, a spotted seatrout must be between 10 and 30 inches long before it can be kept; otherwis

Mademuasel [1]

Answer:

There is a 97.585% probability that a fisherman catches a spotted seatrout within the legal limits.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that:

The lengths of the spotted seatrout that are caught, are normally distributed with a mean of 22 inches, and a standard deviation of 4 inches, so $\mu = 22, \sigma = 4$ .