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3 years ago

On my math test it asks "multiply (-7)(2)(-2)

Mathematics

2 answers:

denis-greek [22]3 years ago

5 0

Answer:

-7*2*-2=28

Step-by-step explanation:

-7*2=-14

-14*-2=28

Hope this helps

STatiana [176]3 years ago

4 0

Answer:

Step-by-step explanation:

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Suppose that a brand of lightbulb lasts on average 1730 hours with a standard deviation of 257 hours. Assume the life of the lig

Debora [2.8K]

Answer:

P [ 1689 ≤ X ≤ 2267 ] = 54,88 %

Step-by-step explanation:

Normal Distribution

Mean μ₀ = 1730

Standard Deviation σ = 257

We need to calculate z scores for the values 1689 and 2267

We apply formula for z scores

z = ( X - μ₀ ) /σ

X = 1689 then

z = (1689 - 1730)/ 257 ⇒ z = - 41 / 257

z = - 0.1595

And from z table we get for z = - 0,1595

We have to interpolate

- 0,15 0,4364

- 0,16 0,4325

Δ = 0.01 0.0039

0,1595 - 0,15 = 0.0095

By rule of three

0,01 0,0039

0,0095 x ?? x = 0.0037

And 0,4364 - 0.0037 = 0,4327

Then P [ X ≤ 1689 ] = 0.4327 or P [ X ≤ 1689 ] = 43,27 %

And for the upper limit 2267 z score will be

z = ( X - 1730 ) / 257 ⇒ z = 537 / 257

z = 2.0894

Now from z table we find for score 2.0894

We interpolate and assume 0.9815

P [ X ≤ 2267 ] = 0,9815

Ths vale already contains th value of P [ X ≤ 1689 ] = 0.4327

Then we subtract to get 0,9815 - 0,4327 = 0,5488

Finally

P [ 1689 ≤ X ≤ 2267 ] = 0,5488 or P [ 1689 ≤ X ≤ 2267 ] = 54,88 %

8 0

3 years ago

Rewrite the expressions in each pair so that they have the same base. c) (1 /2)^2x and (1/ 4)^x - 1

just olya [345]

I'm not completely sure but this is what I would do.
evaluate <span>(1/ 4)^x - 1 </span>as is. But change the (1 /2)^2x to (2/4)^2x. This way both fractions have the same denominator and in this sense, the same base. The 2/4 base still evaluates into 1/2 so nothing, mathematically, is being broken here.

7 0

4 years ago

. Could someone please explain this to me in detail. I am very confused. Thank you so much!. . A candle is 17 inches tall after

andrew-mc [135]

The generic equation of the line is:

$H-H0 = m (T-T0)$