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azamat
4 years ago
15

A local hamburger shop sold a combined total of 763 hamburgers and cheeseburgers on Saturday. There were 63 more cheeseburgers s

old than hamburgers. How many hamburgers were sold on Saturday?
Mathematics
1 answer:
Airida [17]4 years ago
6 0
In order to solve this we'll start by assigning variables to hamburgers and cheeseburgers, since these are what we're trying to find. Lets say x = hamburgers and y = cheeseburgers. So we know two things, we know that x+y= 763 (hamburgers plus cheeseburgers sold equals 763, and we know that y= x+63 (cheeseburgers sold equals 63 more than hamburgers sold). Now we have a system of equations. This can be solved most easily by rearranging each equation to each y, and then set them equal to each other:
x+y=763 -> y=763-x, and we already have y=x+63. Set them equal to each other:
x+63 = 763-x (add x to both sides) -> 2x+63 = 763 (subtract 63 from both sides) -> 2x = 700 (divide both sides by 2) x = 350. So we solved for x, which is hamburgers sold, which is what the question asks for, so your answer is 350 hamburgers were sold on Saturday
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Q=rational number=1/6

2/3+q=1/2=1/6

4 0
3 years ago
A graduate school entrance exam has scores that are normally distributed with a mean of 560 and a standard deviation of 90. What
9966 [12]

26.8% of examinees will score between 600 and 700.

This question is based on z score concept.

A Z-score is a numerical measurement that describes a value's relationship to the mean of a group of values. Z-score is measured in terms of standard deviations from the mean. If a Z-score is 0, it indicates that the data point's score is identical to the mean score.

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

where:

μ is the mean

σ is the standard deviation of the population

Given:

μ = 560

σ = 90

For

600≤ X≤700

for x = 700

Z score =x - μ/σ

=(700 - 560)/90

      = 1.55556

P-value from Z-Table:

P(560<x<700) = P(x<700) - 0.5 = 0.44009

for x = 600

Z score =x - μ/σ

=(600 - 560)/90

      = 0.44444

P-value from Z-Table:

P(560<x<600) = P(x<600) - 0.5 = 0.17164

∴ P(600<x<700) = P(560<x<700) - P(560<x<600)

                           = 0.44009 - 0.17164

                          =0.26845

∴26.8% percentage of examinees will score between 600 and 700.

Learn more about Z score here :

brainly.com/question/15016913

#SPJ4

4 0
2 years ago
Any suggestions or answers
OLEGan [10]
2nd answer.

any pts in the shaded region satisfies your simultaneous equation.
5 0
4 years ago
In a football or soccer game, you have 22 players, from both teams, in the field. what is the probability of having at least any
Tamiku [17]

We can solve this problem using complementary events. Two events are said to be complementary if one negates the other, i.e. E and F are complementary if

E \cap F = \emptyset,\quad E \cup F = \Omega

where \Omega is the whole sample space.

This implies that

P(E) + P(F) = P(\Omega)=1 \implies P(E) = 1-P(F)

So, let's compute the probability that all 22 footballer were born on different days.

The first footballer can be born on any day, since we have no restrictions so far. Since we're using numbers from 1 to 365 to represent days, let's say that the first footballer was born on the day d_1.

The second footballer can be born on any other day, so he has 364 possible birthdays:

d_2 \in \{1,2,3,\ldots 365\} \setminus \{d_1\}

the probability for the first two footballers to be born on two different days is thus

1 \cdot \dfrac{364}{365} = \dfrac{364}{365}

Similarly, the third footballer can be born on any day, except d_1 and d_2:

d_3 \in \{1,2,3,\ldots 365\} \setminus \{d_1,d_2\}

so, the probability for the first three footballers to be born on three different days is

1 \cdot \dfrac{364}{365} \cdot \dfrac{363}{365}

And so on. With each new footballer we include, we have less and less options out of the 365 days, since more and more days will be already occupied by another footballer, and we can't have two players born on the same day.

The probability of all 22 footballers being born on 22 different days is thus

\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

So, the probability that at least two footballers are born on the same day is

1-\dfrac{364\cdot 363 \cdot \ldots \cdot (365-21)}{365^{21}}

since the two events are complementary.

8 0
3 years ago
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6 0
3 years ago
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