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2 years ago

13

Question 6 (4 points)

2 answers:

Kay [80]2 years ago

3 0

Answer:

I believe it would be answer 2)

Step-by-step explanation:

The team has 4 players and the line graph starts at 5 increasing to 9

choli [55]2 years ago

3 0

Answer:

Answer 2

Step-by-step explanation:

They need 5 or more players.

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What is the sum of the geometric sequence 1,3,9... if there were 11 terms?

Delicious77 [7]

Answer:

88,573

Step-by-step explanation:

3 0

2 years ago

You have $83 in your bank account each week you plan to deposit eight dollars from your allowance and $15 from your paycheck the

docker41 [41]

We know that you have 83$ in your bank account.

Okay so what is 15+8=23 So that means you get 23$ per week.

Now what is 83+23=106
106+23=129
129+23=152
152+23=175
Now that was four 23's

That means it would take 4 weeks for you to have 175 dollars in your bank account.
Hope this helped.

7 0

2 years ago

In the diagram, the measures of <1, <4, and <5 are 120°. The measure of <7

RSB [31]

Answer:

yes because angle 1 and 5 are congruent

6 0

2 years ago

Angle A is 5x-15. Angle B is 2x+21. Solve for x then find the measure for Angle B

Komok [63]

Are A and B supplementary or complementary

3 0

2 years ago

Historical data for a local manufacturing company show that the average number of defects per product produced is 2. In addition

charle [14.2K]

Answer:

The probability that there will be a total of 7 defects on four units is 0.14.

Step-by-step explanation:

A Poisson distribution describes the probability distribution of number of success in a specified time interval.

The probability distribution function for a Poisson distribution is:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}, x=0,1,2,3,...$

Let <em>X</em> = number of defects in a unit produced.

It is provided that there are, on average, 2 defects per unit produced.

Then in 4 units the number of defects is, $(2\times4)=8$ .

Compute the probability of exactly 7 defects in 4 units as follows:

$P(X = x)=\frac{e^{-\lambda}\lambda^{x}}{x!}\\P(X=7)=\frac{e^{-8}8^{7}}{7!}\\=\frac{0.0003355\times2097152}{5040}\\ =0.1396\\\approx0.14$

Thus, the probability of exactly 7 defects in 4 units is 0.14.

8 0

2 years ago