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3 years ago

Find the distance between the points (4,-3) and (5,1) Distance =

Mathematics

1 answer:

Serggg [28]3 years ago

6 0

Answer:

d = sqrt(17)

Step-by-step explanation:

d = sqrt( x2 - x1)^2 + (y2 - y1)^2 )

d = sqrt( (4 - 5)^2 + (-3 - 1)^2 )

d = sqrt( (-1)^2 + (-4)^2 )

d = sqrt ( 1 + 16)

d = sqrt(17)

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A bag contains 10 red marbles, 7 green marbles. and 8 striped marbles. A marble is picked, then put back into bag. Find P(green

SIZIF [17.4K]

G = number of green = 7

S = number of striped = 8

T = number total = 10+7+8 = 25

probability of picking green = P(G) = G/T = 7/25

probability of picking striped = P(S) = S/T = 8/25

P(green and striped) = P(G)*P(S) ... events are independent

P(green and striped) = (7/25)*(8/25)

P(green and striped) = (7*8)/(25*25)

P(green and striped) = 56/625

P(green and striped) = 0.0896

--------------------------------

In summary, the answer as a fraction is 56/625

In decimal form, the answer is 0.0896

The value 0.0896 can be converted to percent form to get 8.96%

8 0

3 years ago

2 tbsp. of peanut butter have 0.25 of total carbohydrate a day what fraction is that?

sergij07 [2.7K]

0.25 is 1/4 if that's what youre asking, i'm not sure,

6 0

2 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.

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

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

A. Are the expressions 3(m - 2) + 2(m - 2) and 50m - 2) equivalent expressions?

miss Akunina [59]

Answer:

Step-by-step explanation:

3(m-2)+2(m-2)

3m-6+2m-4

3m+2m-6-4

5m-6-4

5m-10

The expressions 3(m-2)+2(m-2) and 50m-2 are not equivalent expressions.

After simplifying the first expression 3(m-2)+2(m-2) in part (a), we got 5m-10, which is different from the second expression, 50m-2.

5 0

2 years ago

Easy cumulative frequency help

nikklg [1K]

Answer:

add the next frequency

Step-by-step explanation:

listen in ur class

3 0

3 years ago