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Reil [10]
3 years ago
6

Adele took a total of 12 quizzes over the course of 2 weeks. After attending 5 weeks of school this quarter, how many quizzes wi

ll Adele have taken in total? Assume the relationship is directly proportional.
?quizzes
Mathematics
2 answers:
xeze [42]3 years ago
8 0

Answer:

30 quizzes

Step-by-step explanation:

12 quizzes in 2 weeks means that there were 6 per week and 6 (quizzes per week) times 5 (number of weeks) would be 30.

Hope this helped!!

Snowcat [4.5K]3 years ago
4 0

Answer:

30, or 42

Step-by-step explanation:

If they took 12 quizzes over the course of 2 weeks, then we can assume that they take 6 quizzes each week (12/2 = 6).

Over the course of 5 weeks, they'll have taken 30 quizzes if the above is true. However, the 'or' is there because I don't know if they're including the past 12 or not. If they are including the 12 quizzes taken over 2 weeks, then the answer is 42. If they are only referring to the 5 weeks of school, then the answer is 30.

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The quarterback of a football team releases a pass at a height of 6 feet above the playing field, and the football is caught by
motikmotik

Answer:

t = 25.1seconds

Step-by-step explanation:

X(t)= 16cos75°t

Y(t)= 6 + (Vosin75°t - 16t^2)

Converting 78 yards to feet :

1 yard = 3 feet

78feet =?

78×3=234 feet

Y(t)= number of feet above the ground at t seconds

VoCos75° t = 234

t = 234/(VoCos75°)

At this time,4 = y(234/VoCos75°)

4 = 6 + VoSin75°(234/VoCos75) - 16(234/VoCos75°)^2

Vo= 140.66ft/s

Time,t= 234/(140.66× Cos75°)

t = 25.087seconds

t= 25.1 seconds (1 decimal place)

8 0
2 years ago
For a certain river, suppose the drought length Y is the number of consecutive time intervals in which the water supply remains
AnnZ [28]

Answer:

a) There is a 9% probability that a drought lasts exactly 3 intervals.

There is an 85.5% probability that a drought lasts at most 3 intervals.

b)There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

Step-by-step explanation:

The geometric distribution is the number of failures expected before you get a success in a series of Bernoulli trials.

It has the following probability density formula:

f(x) = (1-p)^{x}p

In which p is the probability of a success.

The mean of the geometric distribution is given by the following formula:

\mu = \frac{1-p}{p}

The standard deviation of the geometric distribution is given by the following formula:

\sigma = \sqrt{\frac{1-p}{p^{2}}

In this problem, we have that:

p = 0.383

So

\mu = \frac{1-p}{p} = \frac{1-0.383}{0.383} = 1.61

\sigma = \sqrt{\frac{1-p}{p^{2}}} = \sqrt{\frac{1-0.383}{(0.383)^{2}}} = 2.05

(a) What is the probability that a drought lasts exactly 3 intervals?

This is f(3)

f(x) = (1-p)^{x}p

f(3) = (1-0.383)^{3}*(0.383)

f(3) = 0.09

There is a 9% probability that a drought lasts exactly 3 intervals.

At most 3 intervals?

This is P = f(0) + f(1) + f(2) + f(3)

f(x) = (1-p)^{x}p

f(0) = (1-0.383)^{0}*(0.383) = 0.383

f(1) = (1-0.383)^{1}*(0.383) = 0.236

f(2) = (1-0.383)^{2}*(0.383) = 0.146

Previously in this exercise, we found that f(3) = 0.09

So

P = f(0) + f(1) + f(2) + f(3) = 0.383 + 0.236 + 0.146 + 0.09 = 0.855

There is an 85.5% probability that a drought lasts at most 3 intervals.

(b) What is the probability that the length of a drought exceeds its mean value by at least one standard deviation?

This is P(X \geq \mu+\sigma) = P(X \geq 1.61 + 2.05) = P(X \geq 3.66) = P(X \geq 4).

We are working with discrete data, so 3.66 is rounded up to 4.

Either a drought lasts at least four months, or it lasts at most thee. In a), we found that the probability that it lasts at most 3 months is 0.855. The sum of these probabilities is decimal 1. So:

P(X \leq 3) + P(X \geq 4) = 1

0.855 + P(X \geq 4) = 1

P(X \geq 4) = 0.145

There is a 14.5% probability that the length of a drought exceeds its mean value by at least one standard deviation

8 0
3 years ago
Find the area of the following parallelograms
Kay [80]

Answer:

Area of parallelogram: 1/2×h×sum of base lengths

Property of parallelogram: 2 parallel sides

Area = 1/2×4×(9+9)

        = 36 cm^2

3 0
2 years ago
Read 2 more answers
Please, help im being timed !!!
Agata [3.3K]
B (2x+5y=-10) is the answer.
3 0
2 years ago
Let A and B be two events in a sample space S such that
klasskru [66]
P(A|B)<span>P(A intersect B) = 0.2 = P( B intersect A)

</span>A) P(A intersect B) = <span>P(A|B)*P(B)
Replacing the known vallues:
0.2=</span><span>P(A|B)*0.5
Solving for </span><span>P(A|B):
0.2/0.5=</span><span>P(A|B)*0.5/0.5
0.4=</span><span>P(A|B)
</span><span>P(A|B)=0.4
</span>
B) P(B intersect A) = P(B|A)*P(A)
Replacing the known vallues:
0.2=P(B|A)*0.6
Solving for P(B|A):
0.2/0.6=P(B|A)*0.6/0.6
2/6=P(B|A)
1/3=P(B|A)
P(B|A)=1/3
5 0
3 years ago
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