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3 years ago

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I ordered 8 pizzas for my son’s birthday party. Each pizza had 8 slices. If my son and his friends ate 47 slices of pizza at the

party, how many slices will I have leftover?

1 answer:

Andrews [41]3 years ago

6 0

Answer:

17

Step-by-step explanation:

8*8 = 64

64 - 47 = 17

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Algebra Y= -.5 (x-6) squared +18

telo118 [61]

Answer:Its math

Step-by-step explanation:

Its math Its math

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3 years ago

Explain what mistake was made in the problem below and what should be the

romanna [79]

Answer:

Instead of moving the 3x to the opposite side of the equation they moved the 2x. And the correct answer is x=-6 not x=6.

Step-by-step explanation:

Student Work:

2(x + 4) – 8 = 3x+10-4

2x + 8–8 = 3x + 6

2x = 3x + 6

- 3x – 3x

X = 6

My work:

2(x+4)-8=3x+10-4

2x+8-8=3x+6

2x=3x+6

-x=6

x=-6

7 0

3 years ago

What is the value of x?

Angelina_Jolie [31]

Answer:

x = 5.4

Step-by-step explanation:

3/x = 4/7.2

cross-multiply:

4x = 21.6

x = 5.4

5 0

3 years ago

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The m

Elenna [48]

Answer:

Part a: <em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b: <em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c: <em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d: <em>The probability of at least one arrival in a 15-second period is 0.9179.</em>

Step-by-step explanation:

Airline passengers are arriving at an airport independently. The mean arrival rate is 10 passengers per minute. Consider the random variable X to represent the number of passengers arriving per minute. The random variable X follows a Poisson distribution. That is,

$X \sim {\rm{Poisson}}\left( {\lambda = 10} \right)$

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

Substitute the value of λ=10 in the formula as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{{\left( {10} \right)}^x}}}{{x!}}$

Part a:

The probability that there are no arrivals in one minute is calculated by substituting x = 0 in the formula as,

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}}\\\\ = {e^{ - 10}}\\\\ = 0.000045\\\end{array}$

<em>The probability of no arrivals in a one-minute period is 0.000045.</em>

Part b:

The probability mass function of X can be written as,

$P\left( {X = x} \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}};x = 0,1,2, \ldots$

The probability of the arrival of three or fewer passengers in one minute is calculated by substituting \lambda = 10λ=10 and x = 0,1,2,3x=0,1,2,3 in the formula as,

$\begin{array}{c}\\P\left( {X \le 3} \right) = \sum\limits_{x = 0}^3 {\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}} \\\\ = \frac{{{e^{ - 10}}{{\left( {10} \right)}^0}}}{{0!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^1}}}{{1!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^2}}}{{2!}} + \frac{{{e^{ - 10}}{{\left( {10} \right)}^3}}}{{3!}}\\\\ = 0.000045 + 0.00045 + 0.00227 + 0.00756\\\\ = 0.0103\\\end{array}$

<em>The probability of three or fewer passengers arrive in a one-minute period is 0.0103.</em>

Part c:

Consider the random variable Y to denote the passengers arriving in 15 seconds. This means that the random variable Y can be defined as $\frac{X}{4}$

$\begin{array}{c}\\E\left( Y \right) = E\left( {\frac{X}{4}} \right)\\\\ = \frac{1}{4} \times 10\\\\ = 2.5\\\end{array}$

That is,

$Y\sim {\rm{Poisson}}\left( {\lambda = 2.5} \right)$

So, the probability mass function of Y is,

$P\left( {Y = y} \right) = \frac{{{e^{ - \lambda }}{\lambda ^y}}}{{y!}};x = 0,1,2, \ldots$

The probability that there are no arrivals in the 15-second period can be calculated by substituting the value of (λ=2.5) and y as 0 as:

$\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = {e^{ - 2.5}}\\\\ = 0.0821\\\end{array}$

<em>The probability of no arrivals in a 15-second is 0.0821.</em>

Part d:

The probability that there is at least one arrival in a 15-second period is calculated as,

$\begin{array}{c}\\P\left( {X \ge 1} \right) = 1 - P\left( {X < 1} \right)\\\\ = 1 - P\left( {X = 0} \right)\\\\ = 1 - \frac{{{e^{ - 2.5}} \times {{2.5}^0}}}{{0!}}\\\\ = 1 - {e^{ - 2.5}}\\\end{array}$

$\begin{array}{c}\\ = 1 - 0.082\\\\ = 0.9179\\\end{array}$

<em>The probability of at least one arrival in a 15-second period is 0.9179.</em>

7 0

3 years ago

Please check these please

antoniya [11.8K]

-5x=60

The first thing we want to do is to divide -5 on both sides. We do this because -5x, is the same as -5 times x and we want to do the inverse of that.

After we divide on both sides, we get x=60/-5

We know that 60 divided by -5 is -12.

Therefore our answer is: x=-12

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5= n + 2

We always want to get the variable to be alone, so if we subtract by 5 we would be doing the opposite of getting the variable alone. Although, if we subtract by 2 on both sides, we would get the variable alone and the answer would be 3=n

Therefore our answer is: False

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2(x-7)

By looking at this question, we see x being subtracted by 7. Then we see that 2 is being multiplied to that.

Therefore our answer is: C

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The product of a number and 4 increased by 8

In these types of questions, it's smart to break it up. We will start with the product of a number and 4. We know that a number is a variable and if we look at the multiple choice answers we can see that the variable they chose to use is n. We also know that a product is when numbers get multiplied. So that means 4 and n get multiplied which turns into 4n.

For the other part of the problem, we know that increased by 8 means added my 8. Now that we got our answers, we put then together.

Therefore our answer is: A 4n+8

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Pls give a like if this helped you!!!

4 0

2 years ago

Read 2 more answers