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WITCHER [35]
2 years ago
8

Can you please help me I give you extra points ?

Mathematics
2 answers:
brilliants [131]2 years ago
7 0

Answer:

( -1 , -3 ) , ( 4 , -3 ) , (4 , -1 )

kkurt [141]2 years ago
6 0
Answer: 16 centimeters------nice computer by the way

Step-by-step explanation: addition

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Solve 0=2y−8x+10 for y
qaws [65]

Answer:y

y=4x−5

Step-by-step explanation:

3 0
3 years ago
When ordering the Kids' Lunch at Burger Universe, the customer must choose a size, whether or not to have cheese, a type of bun,
AlekseyPX

Answer:

192 Kids' Lunches are possible.

Step-by-step explanation:

The client must make 5 decisions:

1) Size (junior, small, medium, large)

2) Cheese (with or without)

3) Type of bun (white, wheat)

4) Side order (fries, onion rings, fruit cup)

5) Drink (orange, grape, cherry, lemonade)

Since the client must make all 5 decisions all the time, we will multiply the number of options we have:

1) For size, he has 4 options,

2) For cheese, he has 2 options,

3) For type of bun, 2 options,

4) For side order, 3 options,

5) For drink, 4 options.

Therefore, the number of total options he has is 4 x 2 x 2 x 3 x 4 = 192 Kids' Lunches.

6 0
3 years ago
There are 176 freshmen and 18 adults, going on a trip to Great Adventure. The tickets are $60.00 for an individual or $300.00 fo
Sphinxa [80]

Answer: They would choose $300.00 for 10 people

Step-by-step explanation:

It would be cheaper

5 0
3 years ago
Read 2 more answers
Write a real world problem that can be solved using the equation, 5x = 3x + 20
Natali5045456 [20]
-3x from both sides so it looks like 2x=20 and divide by 2 and x=10
8 0
3 years ago
A grocery store’s receipts show that Sunday customer purchases have a skewed distribution with a mean of 27$ and a standard devi
34kurt

Answer:

(a) The probability that the store’s revenues were at least $9,000 is 0.0233.

(b) The revenue of the store on the worst 1% of such days is $7,631.57.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sum of values of X, i.e ∑X, will be approximately normally distributed.  

Then, the mean of the distribution of the sum of values of X is given by,  

 \mu_{X}=n\mu

And the standard deviation of the distribution of the sum of values of X is given by,  

\sigma_{X}=\sqrt{n}\sigma

It is provided that:

\mu=\$27\\\sigma=\$18\\n=310

As the sample size is quite large, i.e. <em>n</em> = 310 > 30, the central limit theorem can be applied to approximate the sampling distribution of the store’s revenues for Sundays by a normal distribution.

(a)

Compute the probability that the store’s revenues were at least $9,000 as follows:

P(S\geq 9000)=P(\frac{S-\mu_{X}}{\sigma_{X}}\geq \frac{9000-(27\times310)}{\sqrt{310}\times 18})\\\\=P(Z\geq 1.99)\\\\=1-P(Z

Thus, the probability that the store’s revenues were at least $9,000 is 0.0233.

(b)

Let <em>s</em> denote the revenue of the store on the worst 1% of such days.

Then, P (S < s) = 0.01.

The corresponding <em>z-</em>value is, -2.33.

Compute the value of <em>s</em> as follows:

z=\frac{s-\mu_{X}}{\sigma_{X}}\\\\-2.33=\frac{s-8370}{316.923}\\\\s=8370-(2.33\times 316.923)\\\\s=7631.56941\\\\s\approx \$7,631.57

Thus, the revenue of the store on the worst 1% of such days is $7,631.57.

5 0
3 years ago
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