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Lena [83]
3 years ago
15

(02.01 LC) Simplify 4 over 7 ÷ 3 over negative 8. negative 32 over 21 negative 3 over 14 3 over 14 32 over 21

Mathematics
1 answer:
zhannawk [14.2K]3 years ago
4 0

Answer:

(4/7)÷ (3/-8). Division of fractions is the same as multiplying their reciprocals. So (4/7) x (-8/3). Now multiply the numerators together and denominators together (make sure you only use the negative sign once). (4 x -8) / (7 x 3) = -32/21

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Solve the following equation. log(3x)=log(2x-4) *
Irina18 [472]

log(3x) = log(2x-4)

taking antilog of both sides:

3x = 2x - 4

3x - 2x = -4                    [subtracting 2x from both sides]

x = -4

and we're done already!

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3 years ago
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2 years ago
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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
Help plz I need the answer real quick???
Fudgin [204]

Answer:

Distance between mall and library is 6.32 miles.

Step-by-step explanation:

Coordinates of the door to the library is at (2, 4) and door to the mall is at (8, 4).

Since, distance between two points (x_1,y_1) and (x_2,y_2) is given by,

Distance = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

By this formula,

Distance between (2, 4) and (8, 4) will be,

Distance = \sqrt{(8-2)^2+(2-4)^2}

               = \sqrt{36+4}

               =  \sqrt{40}

               = 6.32 miles

4 0
2 years ago
There are six multiple-choice questions on an exam, each with four possible answers. (a) Determine the number of possible answer
Lubov Fominskaja [6]

Answer:

Possible\ Answers = 4096

Step-by-step explanation:

Given

Questions = 6

Options = 4 each

Required

Determine the number of possible answers

<em>The first question has 4 possible answers</em>

<em>The second has 4 possible answers</em>

<em>The third has 4 possible answers</em>

<em>The fourth question has 4 possible answers</em>

<em>The fifth has 4 possible answers</em>

<em>The sixth question has 4 possible answers</em>

<em />

Possible\ Answers = 4 * 4 * 4 * 4 * 4 * 4

Possible\ Answers = 4^6

Possible\ Answers = 4096

8 0
3 years ago
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