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Leto [7]
3 years ago
10

The graph below models the value of a $20,000 car t years after it was purchased.

Mathematics
1 answer:
yarga [219]3 years ago
3 0

Answer: B “ Each time, t, is associated with exact one car value, y.

Step-by-step explanation:

I just took the pre-test on edge.

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Laura got a discount of 100 dollars on her purchase of a 5775 bed.What is the percent of decrease?
matrenka [14]
Remember that percent means parts out of 100
x/100

discount is the decrease
so we divide the decrease by the original total and get
100/5775=0.0173
convert to fraction
0.0173/1
make bottom number 100
multiply by 100/100
1.73/100=1.73%
round
2%
5 0
3 years ago
businessText message users receive or send an average of 62.7 text messages per day. How many text messages does a text message
KiRa [710]

Answer:

(a) The probability that a text message user receives or sends three messages per hour is 0.2180.

(b) The probability that a text message user receives or sends more than three messages per hour is 0.2667.

Step-by-step explanation:

Let <em>X</em> = number of text messages receive or send in an hour.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em>.

It is provided that users receive or send 62.7 text messages in 24 hours.

Then the average number of text messages received or sent in an hour is: \lambda=\frac{62.7}{24}= 2.6125.

The probability of a random variable can be computed using the formula:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!} ;\ x=0, 1, 2, 3, ...

(a)

Compute the probability that a text message user receives or sends three messages per hour as follows:

P(X=3)=\frac{e^{-2.6125}(2.6125)^{3}}{3!} =0.21798\approx0.2180

Thus, the probability that a text message user receives or sends three messages per hour is 0.2180.

(b)

Compute the probability that a text message user receives or sends more than three messages per hour as follows:

P (X > 3) = 1 - P (X ≤ 3)

              = 1 - P (X = 0) - P (X = 1) - P (X = 2) - P (X = 3)

             =1-\frac{e^{-2.6125}(2.6125)^{0}}{0!}-\frac{e^{-2.6125}(2.6125)^{1}}{1!}-\frac{e^{-2.6125}(2.6125)^{2}}{2!}-\frac{e^{-2.6125}(2.6125)^{3}}{3!}\\=1-0.0734-0.1916-0.2503-0.2180\\=0.2667

Thus, the probability that a text message user receives or sends more than three messages per hour is 0.2667.

6 0
3 years ago
Need help on this!!!!
kotykmax [81]

Answer:

D

Step-by-step explanation:

First find h(3)

2(3)-2= 4

then find g(4)

4³-5= 59

4 0
3 years ago
Read 2 more answers
The ratio of the measures of the three angles in a triangle is 9:3:6.
Alex

Step-by-step explanation:

this is how you solve it :)

5 0
2 years ago
A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 434 gram setting. It is
yan [13]

Answer:

No, there is not sufficient evidence to support the claim that the bags are underfilled

Step-by-step explanation:

A manufacturer of potato chips would like to know whether its bag filling machine works correctly at the 434 gram setting.

This means that the null hypothesis is:

H_{0}: \mu = 434

It is believed that the machine is underfilling the bags.

This means that the alternate hypothesis is:

H_{a}: \mu < 434

The test statistic is:

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

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

434 is tested at the null hypothesis:

This means that \mu = 434

A 9 bag sample had a mean of 431 grams with a variance of 144.

This means that X = 431, n = 9, \sigma = \sqrt{144} = 12

Value of the test-statistic:

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

z = \frac{431 - 434}{\frac{12}{\sqrt{9}}}

z = -0.75

P-value of the test:

The pvalue of the test is the pvalue of z = -0.75, which is 0.2266

0.2266 > 0.01, which means that there is not sufficient evidence to support the claim that the bags are underfilled.

7 0
2 years ago
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