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3 years ago

When 508,000,000 is written in scientific notation, what will be the value of the decimal number?

Mathematics

1 answer:

Masja [62]3 years ago

7 0

It would be 5.08x10^8

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1 2 3 4 5 6 7 8 9 10 TIME REMAINING 54:32 Because the lid of a marker is wider than the marker itself, a set of markers can be p

tia_tia [17]

Answer:

10 Square Inches

Step-by-step explanation:

Trapezoid

Base =12 inches

Height =10 inches

Top side= 10 inches.

Area of a Trapezoid $=\frac{1}{2}(a+b)h, $where a and b are the lengths of the base and top respectively$$

$=\frac{1}{2}(10+12)*10\\=5*22=110 \:Square\:Inches$

Area of the Trapezoid=110 Square Inches

Rectangle

Base = 12 inches

Height =10 inches.

Area of a Rectangle=Base X Height

=12 X 10

=120 Square Inches

Difference in Area between the two packages

Difference=Area of Rectangle-Area of Trapezoid

=120-110

=10 Square Inches

5 0

3 years ago

Read 2 more answers

This is the table i need help with

Papessa [141]

2.3 is the other length
Hope this helps :)

8 0

3 years ago

Read 2 more answers

Which one is a better deal?

saul85 [17]

12 pack is a better deal

5 0

3 years ago

? 0.07 What is 70 x 10^2

Fofino [41]

Answer:

7000

Step-by-step explanation:

I think this is the answer

4 0

3 years ago

(a) Find the size of each of two samples (assume that they are of equal size) needed to estimate the difference between the prop

zalisa [80]

Answer:

(a) The sample sizes are 6787.

(b) The sample sizes are 6666.

Step-by-step explanation:

(a)

The information provided is:

Confidence level = 98%

MOE = 0.02

n₁ = n₂ = n

$\hat p_{1} = \hat p_{2} = \hat p = 0.50\ (\text{Assume})$

Compute the sample sizes as follows:

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

$n=\frac{2\times\hat p(1-\hat p)\times (z_{\alpha/2})^{2}}{MOE^{2}}$

$=\frac{2\times0.50(1-0.50)\times (2.33)^{2}}{0.02^{2}}\\\\=6786.125\\\\\approx 6787$

Thus, the sample sizes are 6787.

(b)

Now it is provided that:

$\hat p_{1}=0.45\\\hat p_{2}=0.58$

Compute the sample size as follows:

$MOE=z_{\alpha/2}\times\sqrt{\frac{\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})}{n}$

$n=\frac{(z_{\alpha/2})^{2}\times [\hat p_{1}(1-\hat p_{1})+\hat p_{2}(1-\hat p_{2})]}{MOE^{2}}$

$=\frac{2.33^{2}\times [0.45(1-0.45)+0.58(1-0.58)]}{0.02^{2}}\\\\=6665.331975\\\\\approx 6666$

Thus, the sample sizes are 6666.

7 0

3 years ago