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2 years ago

Can someone help me with #6?

Mathematics

1 answer:

lisov135 [29]2 years ago

8 0

F(x)=4|x+(1/2)|

That's the answer

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Find the product of (5.2 · 10^-6) and (8 · 10^3).

bezimeni [28]

$\displaystyle\\(5.2\cdot10^{-6})\times(8\cdot10^3)=\\\\=\underbrace{5.2\times8}_{41.6}\times\underbrace{10^{-6}\times10^{3}}_{10^{-6+3}}=\\\\=41.6\times10^{-6+3}=\boxed{\bf41.6\times10^{-3}}\\\\\texttt{Correct answer:}~~\boxed{\bf C)}$

7 0

2 years ago

Read 2 more answers

Nicholas volunteers at a local soup kitchen and he notices how quickly the kitchen is going through its stock of canned goods. B

kkurt [141]

Answer:

m=-63 b=825

Step-by-step explanation:

To get m, make sure you use rise over run. y2-y1 over x2-x1

8 0

2 years ago

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A Write an explicit formula for an, the nth term of the sequence 17, 15, 13, ...

yawa3891 [41]

Tn=-2n+19

Step-by-step explanation:

Well in south africa we use Tn

Tn=a+(n-1)d

a=is the first term in the sequence which is 17

d=is the the difference between the terms which is - 2

Tn=17+(n-1)(-2)

Tn=17-2n+2

Tn=-2n+19

3 0

2 years ago

I need help, please. it would mean a lot I need it now

fomenos

Answer:

60 feet

Step-by-step explanation:

a - 2x

b - x

c - 3x

120/6 = 20

20 x 3 = 60

8 0

3 years ago

Read 2 more answers

High-power experimental engines are being developed by the Stevens Motor Company for use in its new sports coupe. The engineers

Maurinko [17]

Answer:

The 99% of a confidence interval for the average maximum HP for the experimental engine.

(536.46, 603.54)

Step-by-step explanation:

Step:-1

Given that the mean of the Population = 540HP

Given that the size of the sample 'n' = 9

Given that the mean of the sample = 570HP

Given that the sample standard deviation = 30HP

Step(ii):-

Degrees of freedom = n-1 =9-1 =8

t₀.₀₀₅ = 3.3554

The 99% of a confidence interval for the average maximum HP for the experimental engine.

$(x^{-} - t_{\frac{0.01}{2} ,8} \frac{S.D}{\sqrt{n} } ,x^{2} + t_{\frac{0.01}{2},8 } \frac{S.D}{\sqrt{n} } )$

$(570-3.354\frac{30}{\sqrt{9} } , 570+3.354\frac{30}{\sqrt{9} } )$

(570 - 33.54 , 570+33.54)

(536.46 , 603.54)

Final answer :-

The 99% of a confidence interval for the average maximum HP for the experimental engine.

(536.46, 603.54)

4 0

2 years ago