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3 years ago

Solve -5(x-7)^2+30=-20

Mathematics

1 answer:

Alex Ar [27]3 years ago

8 0

Answer:

x=7−√10 or x=7+√10

Step-by-step explanation:

−5(x−7)2+30=−20

Step 1: Simplify both sides of the equation.

−5x2+70x−215=−20

Step 2: Subtract -20 from both sides.

−5x2+70x−215−(−20)=−20−(−20)

−5x2+70x−195=0

For this equation: a=-5, b=70, c=-195

−5x2+70x+−195=0

Step 3: Use quadratic formula with a=-5, b=70, c=-195.

−b±√b2−4ac

−(70)±√(70)2−4(−5)(−195)

2(−5)

−70±√1000

−10

x=7−√10 or x=7+√10

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Find the mean, median, mode, and range of the data below:

Inga [223]

Answer:

6.75

Step-by-step explanation:

Find the mean of the set {2,5,5,6,8,8,9,11} .

There are 8 numbers in the set. Add them all, and then divide by 8 .

2 + 5 + 5 + 6 + 8 + 8 + 9 + 118=548=6.75

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3 years ago

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.

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