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Bogdan [553]
2 years ago
11

Find the zeros of the polynomial function and state the multiplicity of each

Mathematics
1 answer:
Murrr4er [49]2 years ago
6 0
Hi the answer zero ther u go
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A coin is tossed and a number cube is rolled at the same time. What is the probability that the coin will land on tails and the
babymother [125]
A-1/12 because if you write down h1, h2, h3, etc.-same for tail….tail and four would show up once out of the twelve options.
8 0
2 years ago
Tamara paid $3.00 for cartons of milk. Teo paid $3.75 for a different number of cartons at the same store. How much is one carto
Rashid [163]
1- Price of each carton:
Since they both bought different number of cartons of milk from the same store, we can simply find the price of one carton from the difference between what they both paid.

Since Tamara paid 3$ and Toe paid 3.75$, we can simply say that one carton of milk costs 0.75$ and that Toe bought one carton more than Tamra.

2- How many cartons each bought:
Tamra bought : 3 / 0.75 = 4 cartons
Toe bought : 3.75 / 0.75 = 5 cartons
3 0
3 years ago
Which point best approximates ? <br><br><br> A<br> B<br> C<br> D
snow_lady [41]

√36 = 6

√49 = 7

√45 is between 6 and 7


so √45 must be less than 7 but close to 7


Estimate answer is C.

6 0
3 years ago
Read 2 more answers
Suppose that 40 percent of the drivers stopped at State Police checkpoints in Storrs on Spring Weekend show evidence of driving
lesantik [10]

Answer:

a) 0.778

b) 0.9222

c) 0.6826

d) 0.3174

e) 2 drivers

Step-by-step explanation:

Given:

Sample size, n = 5

P = 40% = 0.4

a) Probability that none of the drivers shows evidence of intoxication.

P(x=0) = ^nC_x P^x (1-P)^n^-^x

P(x=0) = ^5C_0  (0.4)^0 (1-0.4)^5^-^0

P(x=0) = ^5C_0 (0.4)^0 (0.60)^5

P(x=0) = 0.778

b) Probability that at least one of the drivers shows evidence of intoxication would be:

P(X ≥ 1) = 1 - P(X < 1)

= 1 - P(X = 0)

= 1 - ^5C_0 (0.4)^0 * (0.6)^5

= 1 - 0.0778

= 0.9222

c) The probability that at most two of the drivers show evidence of intoxication.

P(x≤2) = P(X = 0) + P(X = 1) + P(X = 2)

^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2  (0.4)^2  (0.6)^3

= 0.6826

d) Probability that more than two of the drivers show evidence of intoxication.

P(x>2) = 1 - P(X ≤ 2)

= 1 - [^5C_0  (0.4)^0  (0.6)^5 + ^5C_1  (0.4)^1  (0.6)^4 + ^5C_2 * (0.4)^2  (0.6)^3]

= 1 - 0.6826

= 0.3174

e) Expected number of intoxicated drivers.

To find this, use:

Sample size multiplied by sample proportion

n * p

= 5 * 0.40

= 2

Expected number of intoxicated drivers would be 2

7 0
3 years ago
Pls say hard very hard​
Simora [160]

(a) From the histogram, you can see that there are 2 students with scores between 50 and 60; 3 between 60 and 70; 7 between 70 and 80; 9 between 80 and 90; and 1 between 90 and 100. So there are a total of 2 + 3 + 7 + 9 + 1 = 22 students.

(b) This is entirely up to whoever constructed the histogram to begin with... It's ambiguous as to which of the groups contains students with a score of exactly 60 - are they placed in the 50-60 group, or in the 60-70 group?

On the other hand, if a student gets a score of 100, then they would certainly be put in the 90-100 group. So for the sake of consistency, you should probably assume that the groups are assigned as follows:

50 ≤ score ≤ 60   ==>   50-60

60 < score ≤ 70   ==>   60-70

70 < score ≤ 80   ==>   70-80

80 < score ≤ 90   ==>   80-90

90 < score ≤ 100   ==>   90-100

Then a student who scored a 60 should be added to the 50-60 category.

8 0
2 years ago
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