Estimate each quotient 53.4÷6.15

When multiplying 4.73 by 2.1 how many decimal places will the product have

hichkok12 [17]

it would be 9.93300

so 3 decimal places because the zeros don't count

3 0

4 years ago

Read 2 more answers

Sovle the following inequality -3m + 18 < 30

denis-greek [22]

Answer:

m>-4

Step-by-step explanation: First subtract 18 from both sides, then simplify 30-18, which equals 12. Divide both sides by -3, and flip your sign.

7 0

3 years ago

Can anyone help me with this one :D

allsm [11]

Answer:

the answers is (10^1)^3 = 1000

4 0

3 years ago

Mikaela placed a frame around a print that measures 10 inches by 10 inches. The area of just the frame itself is 69 square inche

olchik [2.2K]

Answer:

1.5 in

Step-by-step explanation:

Let x be the width of the frame.

Side of print=10 in

Area of frame=69 square in

We have to find the width of the frame.

Side of frame=10+x+x=10+2x

Area of square= $(side)^2$

By using the formula

Area of print= $10\times 10=100in^2$

Area of frame with print= $(10+2x)^2$

Area of frame=Area of frame with print-Area of print

$69=(10+2x)^2-100$

$(10+2x)^2=69+100=169$

$(10+2x)=\sqrt{169}=\pm 13$

$10+2x=13$

Because Side is always positive.

$2x=13-10=3$

$x=\frac{3}{2}=1.5$

Hence, width of frame=1.5 in

3 0

3 years ago

Suppose that you had the following data set. 500 200 250 275 300 Suppose that the value 500 was a typo, and it was suppose to be

hodyreva [135]

Answer:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

Step-by-step explanation:

The subindex B is for the before case and the subindex A is for the after case

Before case (with 500)

For this case we have the following dataset:

500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_B = \frac{\sum_{i=1}^5 X_i}{5} =\frac{500+200+250+275+300}{5}=\frac{1525}{5}=305$

And the sample deviation with the following formula:

$s_B = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(500-305)^2 +(200-305)^2 +(250-305)^2 +(275-305)^2 +(300-305)^2)}{5-1}} = 115.108$

After case (With -500 instead of 500)

For this case we have the following dataset:

-500 200 250 275 300

We can calculate the mean with the following formula:

$\bar X_A = \frac{\sum_{i=1}^5 X_i}{5} =\frac{-500+200+250+275+300}{5}=\frac{525}{5}=105$

And the sample deviation with the following formula:

$s_A = \sqrt{\frac{\sum_{i=1}^5 (X_i-\bar X)^2}{n-1}}=\sqrt{\frac{(-500-105)^2 +(200-105)^2 +(250-105)^2 +(275-105)^2 +(300-105)^2)}{5-1}} = 340.221$

And as we can see we have a significant change between the two values for the two cases.

The absolute difference is:

$Abs = |340.221-115.108|= 225.113$

If we find the % of change respect the before case we have this:

$\% Change = \frac{|340.221-115.108|}{115.108} *100 = 195.57\%$

So then is a big change.

8 0

4 years ago