Which function represents the amount of money Melanie and Denzel have left, f(x), after buying x pounds of strawberries?

An elevator went up 15 floors, down 9 floors, up 11 floors, and down 19 floors. Find the net change.

NARA [144]

Answer: The net change would be -2.

Step-by-step explanation: The upward change of elevator is represented by positive sign while the downward change represented by negative sign.

Given, First the elevator goes up 15 floors, change = + 15, Then it goes 9 floors down, change = -9, Then it goes 11 floors up,Change = + 11, Finally, it goes 19 floor down, Change = - 19 So, the net change = sum of total changes = 15 - 9 + 11 - 19 = -2

your welcome!

6 0

2 years ago

Rain fell on four days in June. These amounts were recorded: 5 mm, 9 mm, 2 cm, and 2.6 cm. What was the total rainfall for June?

ivolga24 [154]

Hello there

The correct answer in mm would be 60 mm

1 cm = 10 mm

So it would 5+9+20+26 = 60

In cm it would be 6 cm

10 mm = 1 cm

.5+.9+2+2.6 = 6

I hope this helped ^^

8 0

3 years ago

Dr. Miriam Johnson has been teaching accounting for over 20 years. From her experience, she knows that 60% of her students do ho

oksano4ka [1.4K]

Answer:

a) The probability that a student will do homework regularly and also pass the course = P(H n P) = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P') = 0.12

c) The two events, pass the course and do homework regularly, aren't mutually exclusive. Check Explanation for reasons why.

d) The two events, pass the course and do homework regularly, aren't independent. Check Explanation for reasons why.

Step-by-step explanation:

Let the event that a student does homework regularly be H.

The event that a student passes the course be P.

- 60% of her students do homework regularly

P(H) = 60% = 0.60

- 95% of the students who do their homework regularly generally pass the course

P(P|H) = 95% = 0.95

- She also knows that 85% of her students pass the course.

P(P) = 85% = 0.85

a) The probability that a student will do homework regularly and also pass the course = P(H n P)

The conditional probability of A occurring given that B has occurred, P(A|B), is given as

P(A|B) = P(A n B) ÷ P(B)

And we can write that

P(A n B) = P(A|B) × P(B)

Hence,

P(H n P) = P(P n H) = P(P|H) × P(H) = 0.95 × 0.60 = 0.57

b) The probability that a student will neither do homework regularly nor will pass the course = P(H' n P')

From Sets Theory,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

P(H n P) = 0.57 (from (a))

Note also that

P(H) = P(H n P') + P(H n P) (since the events P and P' are mutually exclusive)

0.60 = P(H n P') + 0.57

P(H n P') = 0.60 - 0.57

Also

P(P) = P(H' n P) + P(H n P) (since the events H and H' are mutually exclusive)

0.85 = P(H' n P) + 0.57

P(H' n P) = 0.85 - 0.57 = 0.28

So,

P(H n P') + P(H' n P) + P(H n P) + P(H' n P') = 1

Becomes

0.03 + 0.28 + 0.57 + P(H' n P') = 1

P(H' n P') = 1 - 0.03 - 0.57 - 0.28 = 0.12

c) Are the events "pass the course" and "do homework regularly" mutually exclusive? Explain.

Two events are said to be mutually exclusive if the two events cannot take place at the same time. The mathematical statement used to confirm the mutual exclusivity of two events A and B is that if A and B are mutually exclusive,

P(A n B) = 0.

But, P(H n P) has been calculated to be 0.57, P(H n P) = 0.57 ≠ 0.

Hence, the two events aren't mutually exclusive.

d. Are the events "pass the course" and "do homework regularly" independent? Explain

Two events are said to be independent of the probabilty of one occurring dowant depend on the probability of the other one occurring. It sis proven mathematically that two events A and B are independent when

P(A|B) = P(A)

P(B|A) = P(B)

P(A n B) = P(A) × P(B)

To check if the events pass the course and do homework regularly are mutually exclusive now.

P(P|H) = 0.95

P(P) = 0.85

P(H|P) = P(P n H) ÷ P(P) = 0.57 ÷ 0.85 = 0.671

P(H) = 0.60

P(H n P) = P(P n H)

P(P|H) = 0.95 ≠ 0.85 = P(P)

P(H|P) = 0.671 ≠ 0.60 = P(H)

P(P)×P(H) = 0.85 × 0.60 = 0.51 ≠ 0.57 = P(P n H)

None of the conditions is satisfied, hence, we can conclude that the two events are not independent.

Hope this Helps!!!

7 0

3 years ago

The probability of rain on the last day of July is 95 % . If the probability remains constant for the first seven days of August

8090 [49]

The probability that it rains at most 2 days is 0.00005995233 and the variance is 0.516

<h3>The probability that it rains at most 2 days</h3>

The given parameters are:

Number of days, n = 7
Probability that it rains, p = 95%
Number of days it rains, x = 2 (at most)

The probability that it rains at most 2 days is represented as:

P(x ≤ 2) = P(0) + P(1) + P(2)

Each probability is calculated as:

$P(x) = ^nC_x * p^x * (1 - p)^{n - x}$

So, we have:

$P(0) = ^7C_0 * (92\%)^0 * (1 - 92\%)^{7 - 0} = 0.00000002097$

$P(1) = ^7C_1 * (92\%)^1 * (1 - 92\%)^{7 - 1} = 0.00000168821$

$P(2) = ^7C_2 * (92\%)^2 * (1 - 92\%)^{7 - 2} = 0.00005824315$

So, we have:

P(x ≤ 2) =0.00000002097 + 0.00000168821 + 0.00005824315

P(x ≤ 2) = 0.00005995233

Hence, the probability that it rains at most 2 days is 0.00005995233

This is calculated as:

Mean = np

So, we have:

Mean = 7 * 92%

Evaluate

Mean = 6.44

Hence, the mean is 6.44

<h3>The standard deviation</h3>

This is calculated as:

σ = √np(1 - p)

So, we have:

σ = √7 * 92%(1 - 92%)

Evaluate

σ = 0.718

Hence, the standard deviation is 0.718

<h3>The variance</h3>

We have:

σ = 0.718

Square both sides

σ² = 0.718²

Evaluate

σ² = 0.516

This represents the variance

Hence, the variance is 0.516

Read more about normal distribution at:

brainly.com/question/4079902

#SPJ1

7 0

1 year ago

Can someone PLEASE help me this is so hardddddd!!!!!

vfiekz [6]

Answer:

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8 0

2 years ago