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xeze [42]
3 years ago
12

Find the distance between -508 and 123.​

Mathematics
2 answers:
Vikentia [17]3 years ago
7 0
Pretty sure the answer is 631
Luda [366]3 years ago
3 0

Answer:

add the absolute value for both, boom ur answer

Step-by-step explanation:

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If Breanna walked 3 1/4 kilometers each day for 12 days, what was the total distance that she walked?
romanna [79]

Answer:

C 39 km

Step-by-step explanation:

Just do 3.25 (3 and 1/4) times 12 and get 39.

7 0
3 years ago
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You have $190 to buy 8 model trains for
lubasha [3.4K]

Answer:

8x; x ≤ $23.75

Step-by-step explanation:

You have to buy 8 trains with 190 dollars. You can't use more money or have more trains.

8x ≤ 190

Divide by 8 on both sides.

x ≤ $23.75 per train

Your welcome!

Kayden Kohl

8th Grade Student

6 0
2 years ago
Let f(x) = 3x + 5 and g(x) = -4x + 7. Find.(f.g)(-4)
Shalnov [3]

Answer:

74

Step-by-step explanation:

fg(x)=f[g(x)]

=f(-4x+7)

=3(-4x+7)+5

=-12x+26

When x=-4,

fg(-4)=-12(-4)+26

=74

3 0
3 years ago
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Which expression is equivalent to this expression? 4a + (-6b) – 3a + 2b
frutty [35]
First, let’s break down the equation by combining like terms. You get 1a-4b, therefore making it answer choice b.
8 0
2 years ago
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The mean life span of a brand name tire is 50,000 miles. Assume that the life spans of the tires are normally distributed, and t
Vlada [557]

Answer:

a) P(X

b) P(\bar X>50200)=1-0.994=0.0062  

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Let X the random variable that represent the mean life span of a brand name tire, and for this case we know the distribution for X is given by:

X \sim N(\mu=50000,\sigma=800)  

Part a

We want this probability:

P(X

The best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

P(X

Part b

Let \bar X represent the sample mean, the distribution for the sample mean is given by:

\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})

On this case  \bar X \sim N(50000,\frac{800}{\sqrt{100}})

We want this probability:

P(\bar X>50200)=1-P(\bar X

The best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}

If we apply this formula to our probability we got this:

P(\bar X >50200)=1-P(Z

P(\bar X>50200)=1-0.994=0.0062  

3 0
3 years ago
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