Find an equation of the line containing the centers of the two circles whose equations are given below.

What is the midpoints of a segment whose endpoints are (-3, 8) and (7, -2) ?

Sidana [21]

$M = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$
$M = (\frac{7 - (-3)}{2}, \frac{-2 + 8}{2})$
$M = (\frac{7 - 3}{2}, \frac{6}{2})$
$M = (\frac{4}{2}, 3)$
$M = (2, 3)$

3 0

3 years ago

Tony made 2 fruit pies. After his family ate some of each pie, 13 of one pie remained and 56 of the other pie remained. About ho

omeli [17]

Answer:

<h2>There remains 7/6 of pie.</h2>

Step-by-step explanation:

Givens

The remaining parts are 1/3 of one pie and 5/6 of the other pie.

To find how much pie is remaining, we just need to sum those fractions.

$\frac{1}{3}+\frac{5}{6}=\frac{6+15}{18} =\frac{21}{18} =\frac{7}{6}$

Therefore, there remains 7/6 of pie. They ate just 1/6.

4 0

3 years ago

Suppose you obtain a chi-square statistic of 67.81. Are your results statistically significant if the critical value obtained fr

Svetach [21]

Answer:

Result is statistically significant.

Step-by-step explanation:

Given that :

Chisquare statistic, χ² = 67.81

Critical value for the distribution, χ²critical = 3.84

α = 0.05

The Decison region :

If χ² statistic > Critical value ; Reject H0 ; this. Eans that result is statistically significant.

Therefore, since, 67.81 > 3.84 ; This means that the result is statistically significant at 0.05

8 0

3 years ago

Let each square represent 1/3 unit by 1/3 unit

7nadin3 [17]

4110000 is the answer

4 0

3 years ago

Read 2 more answers

A certain company sends 40% of its overnight mail parcels by means of express mail service A1. Of these parcels, 4% arrive after

Harrizon [31]

Answer:

(a) The probability that a randomly selected parcel arrived late is 0.026.

(b) The probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c) The probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d) The probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

Step-by-step explanation:

Consider the tree diagram below.

(a)

The law of total probability sates that: $P(A)=P(A|B)P(B)+P(A|B')P(B')$

Use the law of total probability to determine the probability of a parcel being late.

$P(L)=P(L|A_{1})P(A_{1})+P(L|A_{2})P(A_{2})+P(L|A_{3})P(A_{3})\\=(0.04\times0.40)+(0.01\times0.50)+(0.05\times0.10)\\=0.026$

Thus, the probability that a randomly selected parcel arrived late is 0.026.

(b)

The conditional probability of an event A provided that another event B has already occurred is:

$P(A|B)=\frac{P(B|A)P(A)}{P(B)}$

Compute the probability that a parcel was late was being shipped through the overnight mail service A₁ as follows:

$P(A_{1}|L)=\frac{P(L|A_{1})P(A_{1})}{P(L)} \\=\frac{0.04\times 0.40}{0.026} \\=0.615$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₁ is 0.615.

(c)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{2}|L)=\frac{P(L|A_{2})P(A_{2})}{P(L)} \\=\frac{0.01\times 0.50}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₂ is 0.192.

(d)

Compute the probability that a parcel was late was being shipped through the overnight mail service A₂ as follows:

$P(A_{3}|L)=\frac{P(L|A_{3})P(A_{3})}{P(L)} \\=\frac{0.05\times 0.10}{0.026} \\=0.192$

Thus, the probability that a parcel was late was being shipped through the overnight mail service A₃ is 0.192.

4 0

3 years ago