Which is equal to 8^-10/8^5 ?

1 answer:

8 0

$8^{-10}$ can be converted into $\frac{1}{ 8^{10} }$ , and applying this to $8^{-5}$ can see that

$\frac{ \frac{1}{ 8^{10} } }{ \frac{1}{ 8^{5} } }$

This gives us

$\frac{ 8^{5} }{ 8^{10} } = \frac{1}{ 8^{5} } = 8^{-5}$

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What is the equation of the line that passes through the point(−7,−1) and has a slope of 1?

bonufazy [111]

Answer:

y=x+6

Step-by-step explanation:

Put (-7,-1) into

y-y1=m(x-x1) form, and m=1 (the slope)

y+1=1(x+7)

y+1=x+7

y=x+6

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3 years ago

Here are the scores of students on a history test. 55, 56, 57, 58, 58, 64, 65, 66, 69, 75, 77, 78, 81, 81, 85, 86, 91. Create a

Anarel [89]

The first thing you need to do is get the median of the whole set.

The median is the center so since there are 17 numbers the center is 69 and 69 is Q2

Next we need to find Q1 Q3 and Q4 Also, are whiskers are below.
Whisker minimum = 55
Whisker maximum = 91

Now to find Q1 we have to get the median of the first half of the data.
We need to get the median of the following data set.
55, 56, 57, 58, 58, 64, 65, 66

(58+58) / 2 = 58
Q1 = 58

Now we need to find Q3. To do that we get the median of the second half of the data.
75, 77, 78, 81, 81, 85, 86, 91.
(81 + 81) / 2 = 81
Q3 = 81

Now we have all our quartiles and whiskers, which are below, it is time to create our box and whisker plot. Note, create a horizontal or vertical number line with appropriate data. Also the box part is from Q1 to Q3 and your whiskers goes from your minimum to your box and from your maximum to your box.

Whisker Minimum = 55
Q1 = 58
Q2 = 69
Q3 = 81
Whisker Maximum = 91