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3 years ago

A passenger car weighs 4.5 times 10 Superscript 3 units. Which unit of measure was used?

Mathematics

2 answers:

Pepsi [2]3 years ago

5 1

20? I’m not for sure maybe look in the book? That’s the answer I got and it was right

Jay NOOB2 years ago

0 0

Answer is: pounds

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Emily and Sarah had a total of $80. after Sarah spent 1/3 of her money and Emily spent $17, Emily had twice as much money as Sar

Phoenix [80]

<span> SO in total, Emily and Sarah had a total of 80 dollars in which Emily had twice as much as Sarah.
Let’s solve to find out how much their Money is.
=> Since the ratio of the given data is 2:1, 2 + 1 =3, so let’s divide 80 by 3
=> 80 / 3 = 26.667 ,
=> Emily has twice as this.
=> 26.667 * 2 = 53.33
=> Sarah has 26.67
Now, Sarah spent 1/3 of her money
=> 26.67 / 3 = 8.89 – her remaining money
Emily spent 17 dollars of her money
=> 53.33 – 17 = 36.33</span>

3 0

2 years ago

4 to the power of -3 as fraction

shusha [124]

Answer:

Step-by-step explanation:

4^-3

=1/4^3

=1/64

4 0

2 years ago

Read 2 more answers

Simplify 2(x-4)+7(x+4) <br> A. 9x-6 B.9x-4 C.9x+6

MrRissso [65]

2(x - 4) + 7(x + 4)
(2x - 2 • 4) + (7x + 7 • 4)
9x + 20

5 0

2 years ago

Help!

ss7ja [257]

Answer:

1. yes this is proportional

2. your table should include the following numbers

1 = 34

2 = 46

3 = 58

4 = 70

3. 12x + 22 = y

Step-by-step explanation:

5 0

2 years ago

An assembly consists of two mechanical components. Suppose that the probabilities that the first and second components meet spec

harkovskaia [24]

Answer:

P(X = 0) = 0.0208, P(X = 1) = 0.2484, P(X = 2) = 0.7308

Step-by-step explanation:

Let's define the following events

A: the first component meet specification

B: the second component meet specification

P(A) = 0.87

P(B) = 0.84

Let X be the random variable that represents the number of components in the assembly that meet specifications. Because there are only two mechanical components in the assembly, X can only take the values 0, 1, 2.

P(X = 0) = P(the first component does not meet specification and the second component does not meet specification) =

$P(A^{c}\cap B^{c}) = P(A^{c})P(B^{c})$ (because of independence)

= (0.13)(0.16) = 0.0208

P(X = 1) = P(only one component meet specification) = P[(the first component meet specification and the second component does not meet specification) or (the first component does not meet specification and the second component meet specification)] =

$P[(A\cap B^{c})\cup (A^{c}\cap B)] = P(A\cap B^{c}) + P(A^{c}\cap B)=$ (because sets are mutually exclusive)

$P(A)P(B^{c}) + P(A^{c})P(B)=$ (because of independence)

= (0.87)(0.16) + (0.13)(0.84) = 0.2484

P(X = 2) = P(both components meet specifications) =

$P(A\cap B) = P(A)P(B)$ (because of independence)

= (0.87)(0.84)

= 0.7308

8 0

3 years ago