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Ghella [55]
2 years ago
14

Juan delivers newspapers at a daily rate of $30. He also gets an extra $0.25 per newspaper delivered before 8 A.M. If Juan was p

aid $36.25 for Sunday, how many newspapers did he deliver before 8 A.M? *
Mathematics
1 answer:
irina [24]2 years ago
7 0

Answer:

he delivered 25 newspapers be for 8 am

Step-by-step explanation:

have a good day!

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Hep please will give brainiest and other!!!
Vaselesa [24]

y = -9

Ok so to write the equation of the line you will need the slope and y-intercept. Since we are given two-point we can easily find the slope by using this equation for slope formula:

\frac{y_{2}-y_{1}  }{x_{2}-x_{1}  }

So now that we have our equation we can just plug in the numbers:

\frac{(-9)-(-9)}{5-(-3)}

After subtracting you should get:

0/8

Since zero is in the numerator and you can't divide zero by anything, the slope is 0. We still need the y-intercept for the equation however since the slope is 0 there is no need to put anything else.

Then to find the y-intercept all you have to do is plug in one of the coordinatines into the slope equation to solve, for example, using the point (5,-9):

(-9) = 5(0) + b

B is the variable for the y-intercept. Also notice how I put 0 as our slope into the equation. Now all you have to do is solve for b. Which you would get b = -9. Since you have your slope and your y-intercept now you just write out your equatoin for the line which is:

y = 0x - 9

**Just write it as

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Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi
Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

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