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NeX [460]
4 years ago
9

How do you solve this problem? Emma and Anna went to the market to buy 25 fruits. Emma got a few apples that cost $2 each, and A

nna bought some oranges which cost $3 each. If their combined total is $60, how many oranges did Anna get?
Mathematics
1 answer:
sp2606 [1]4 years ago
7 0

Anna bought 10 oranges

Explanation:

Let x denote the number of apples.

Let y denote the number of oranges.

The system of equations can be written as

x+y=25 and 2x+3y=60

Let us solve the equation by substitution method.

Thus, from x+y=25, the value of y can be determined such that y=25-x

Using the y value y=25-x in the equation 2x+3y=60, we get,

2x+3(25-x)=60

Multiplying 3 within the bracket,

2x+75-3x=60

Adding the terms,

75-x=60

Subtracting both sides by 75, we have,

x=15

Thus, substituting x=15 in y=25-x, we get,

y=25-15\\y=10

Thus, Anna get 10 Oranges.

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An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 35, de
Amiraneli [1.4K]

Answer:

a) P=0.1721

b) P=0.3528

c) P=0.3981

Step-by-step explanation:

This sampling can be modeled by a binominal distribution where p is the probability of a project to belong to the first section and q the probability of belonging to the second section.

a) In this case we have a sample size of n=15.

The value of p is p=25/(25+35)=0.4167 and q=1-0.4167=0.5833.

The probability of having exactly 10 projects for the second section is equal to having exactly 5 projects of the first section.

This probability can be calculated as:

P=\frac{n!}{(n-k)!k!}p^kq^{n-k}= \frac{15!}{(10)!5!}\cdot 0.4167^5\cdot0.5833^{10}=0.1721

b) To have at least 10 projects from the 2nd section, means we have at most 5 projects for the first section. In this case, we have to calculate the probability for k=0 (every project belongs to the 2nd section), k=1, k=2, k=3, k=4 and k=5.

We apply the same formula but as a sum:

P(k\leq5)=\sum_{k=0}^{5}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have:

P(k=0)=0.0003\\P(k=1)=0.0033\\P(k=2)=0.0165\\P(k=3)=0.0511\\P(k=4)=0.1095\\P(k=5)=0.1721\\\\P(k\leq5)=0.0003+0.0033+0.0165+0.0511+0.1095+0.1721=0.3528

c) In this case, we have the sum of the probability that k is equal or less than 5, and the probability tha k is 10 or more (10 or more projects belonging to the 1st section).

The first (k less or equal to 5) is already calculated.

We have to calculate for k equal to 10 or more.

P(k\geq10)=\sum_{k=10}^{15}\frac{n!}{(n-k)!k!}p^kq^{n-k}

Then we have

P(k=10)=0.0320\\P(k=11)=0.0104\\P(k=12)=0.0025\\P(k=13)=0.0004\\P(k=14)=0.0000\\P(k=15)=0.0000\\\\P(k\geq10)=0.032+0.0104+0.0025+0.0004+0+0=0.0453

The sum of the probabilities is

P(k\leq5)+P(k\geq10)=0.3528+0.0453=0.3981

8 0
3 years ago
Bill, Lucas, Carmen, and Dena go bowling every week. When ordered from highest to lower, how many ways can their scores be arran
Sauron [17]
20+ answers are out there
3 0
4 years ago
What is the solution to this equation?
pychu [463]

Answer:

A. x = 3

Step-by-step explanation:

5x+15 + 2x = 24 +4x

7x+15=24+4x

7x+15-15=24+4x-15

7x=4x+9

7x-4x=4x+9-4x

3x=9

3x/3=9/3

x=3

4 0
2 years ago
Read 2 more answers
Kateri draws a square. She wants to draw one line to draw one line to divide the square into 2 triangles. Is it possible for her
Vika [28.1K]
No. Since you can only draw from the corners to get triangles, you will end up with two right triangles because of the corners. If it is a right triangle, it can obviously not be obtuse.
6 0
4 years ago
Whoever answers correctly gets 45 points
dimulka [17.4K]

m<1=23

m<2=90

m<3=67

m<4=113

m<5=67

the square means it is 90° so 2 is 90°.

all of the angles combined equals 360 so.

360-67-90= 203

since there is a straight line splitting down the middle. m<4=180-67=113

so m<4=113

now you would go 360-67-113-90=90 so

m<1 + m<3 =90

180-90-67=23.

m<1=23

360-67-23-90-113=67

m<3=67

5 0
3 years ago
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