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3 years ago

-42 divided by (-7) star reading

Mathematics

2 answers:

Ahat [919]3 years ago

5 0

Answer:6

Step-by-step explanation:

madreJ [45]3 years ago

5 0

That would be positive six

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The distance of planet Jupiter from the Sun is approximately 7.8 ⋅ 108 kilometers, and the distance of planet Saturn from the Su

katen-ka-za [31]

Distance of Jupiter to sun:
7.8*108=842.4
Distance of Saturn from sun:
1.5*109=163.5

difference:
842.4-163.5=678.9

6 0

4 years ago

Read 2 more answers

33. Suppose that the scores on a statewide standardized test are normally distributed with a mean of 78 and a standard deviation

tino4ka555 [31]

Given the scores on a statewide standardized test are normally distributed

Mean = μ = 78

Standard deviation = σ = 3

Normalize the data using the z-score by using the following formula and chart:

$z=\frac{x-\mu}{\sigma}$

Estimate the percentage of scores of the following cases:

(a) between 75 and 81

so, the z-score for the given numbers will be:

$\begin{gathered} 75\rightarrow z=\frac{75-78}{3}=\frac{-3}{3}=-1 \\ 81\rightarrow z=\frac{81-78}{3}=\frac{3}{3}=1 \end{gathered}$

As shown, the percentage when (-1 < z < 1) = 68%

(b) above 87

$87\rightarrow z=\frac{87-78}{3}=\frac{9}{3}=3$

The percentage when (z > 3) = 0.5%

(c) below 72

$72\rightarrow z=\frac{72-78}{3}=\frac{-6}{3}=-2$

The percentage when (z < -2) = 0.5 + 2 = 2.5%

(d) between 75 and 84

$\begin{gathered} 75\rightarrow z=\frac{75-78}{3}=-\frac{3}{3}=-1 \\ 84\rightarrow z=\frac{84-78}{3}=\frac{6}{3}=2 \end{gathered}$

The percentage when ( -1 < z < 2 ) = 68 + 13.5 = 81.5%

3 0

1 year ago

How do you write 5 and nine tenths in expanded form

e-lub [12.9K]

90 + 5. if im wrong, comment on my answer i gave you

3 0

4 years ago

Read 2 more answers

If (2i/2+i)-(3i/3+i)=a+bi, then a=<br>A. 1/10<br>B. -10<br>C. 1/50<br>D. -1/10

Verizon [17]

Answer:

Option A is correct.

Step-by-step explanation:

We are given:

$\frac{2i}{2+i}-\frac{3i}{3+i} = a+bi$

We need to find the value of a.

The LCM of (2+i) and (3+i) is (2+i)(3+i)

$=\frac{2i(3+i)}{(2+i)(3+i)}-\frac{3i(2+i)}{(2+i)(3+i)}\\=\frac{6i+2i^2}{(2+i)(3+i)}-\frac{6i+3i^2}{(2+i)(3+i)}\\=\frac{6i+2i^2-(6i+3i^2)}{(2+i)(3+i)}\\=\frac{6i+2i^2-6i-3i^2)}{5+5i}\\=\frac{-i^2}{5+5i}\\i^2=-1\\=\frac{-(-1)}{5+5i}\\=\frac{1}{5+5i}$

Now rationalize the denominator by multiplying by 5-5i/5-5i

$=\frac{1}{5+5i}*\frac{5-5i}{5-5i} \\=\frac{5-5i}{(5+5i)(5-5i)}\\=\frac{5-5i}{(5+5i)(5-5i)}\\(a+b)(a-b)= a^2-b^2\\=\frac{5(1-i)}{(5)^2-(5i)^2}\\=\frac{5(1-i)}{25+25}\\=\frac{5(1-i)}{50}\\=\frac{1-i}{10}\\=\frac{1}{10}-\frac{i}{10}$