Someone help please ://////

Which of the following is the number 5.03 x 10-5 written in standard form?

FromTheMoon [43]

Step-by-step explanation:

Answer=B

proving:

503,000 > 10

so we look for a number

less than 10

=5.03 ×10-5

the other three zeros are not

needed.

Thank you

6 0

3 years ago

Read 2 more answers

On Wednesday, 1/3 of the students wore blue shirts and 1/2 of the students wore white shirts. What fraction of students wore eit

s344n2d4d5 [400]

Should be 5/6 if you balance it by making the denominator the same

5 0

3 years ago

Select the digit in the ten thousand place for 542,970

Vesna [10]

Answer:

the 4

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Factor Trinomials:

notka56 [123]

Answer:

1. $2(x+2)(x+5)$

2. $8(x+4)^{2}$

Step-by-step explanation:

Problem #1:

1. Find the GCF (Greatest Common Factor)

$GCF = 2$

2. Factor out the GCF and simplify.

$2(x^{2} +7x+10)$

3. Factor $x^{2} +7x+10$

Which two numbers add up to 7 and multiply to 10?

2 and 5

Rewrite the expression using the above.

$(x+2)(x+5)$

4. Done!

$2(x+2)(x+5)$

Problem #2:

1. Find the GCF (Greatest Common Factor)

$GCF=8$

2. Factor out the GCF and simplify.

$8(x^{2} +8x+16)$

3. Use the perfect square formula. $a^{2} +2ab+b^{2} =(a+b)^{2}$

$a = x\\b=4$

$(x+4)^{2}$

4. Done!

$8(x+4)^{2}$

8 0

2 years ago

a. Fill in the midpoint of each class in the column provided. b. Enter the midpoints in L1 and the frequencies in L2, and use 1-

Tresset [83]

Answer:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Using the frequency distribution, I found the mean height to be 70.2903 with a standard deviation of 3.5795

Step-by-step explanation:

Given

See attachment for class

Solving (a): Fill the midpoint of each class.

Midpoint (M) is calculated as:

$M = \frac{1}{2}(Lower + Upper)$

Where

$Lower \to$ Lower class interval

$Upper \to$ Upper class interval

So, we have:

Class 63-65:

$M = \frac{1}{2}(63 + 65) = 64$

Class 66 - 68:

$M = \frac{1}{2}(66 + 68) = 67$

When the computation is completed, the frequency distribution will be:

$\begin{array}{ccc}{Midpoint} & {Class} & {Frequency} & {64} & {63-65} & {1} & {67} & {66-68} & {11} & {70} & {69-71} & {8} &{73} & {72-74} & {7} & {76} & {75-77} & {3} & {79} & {78-80} & {1}\ \end{array}$

Solving (b): Mean and standard deviation using 1-VarStats

Using 1-VarStats, the solution is:

$\bar x = 70.2903$

$\sigma = 3.5795$

See attachment for result of 1-VarStats

8 0

3 years ago