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julia-pushkina [17]
2 years ago
14

Martha is planting a garden that will cover up to 400 square feet. She wants to plant two types of flowers, daises and roses. Ea

ch daisy covers 2 square feet and each rose covers 1.5 square feet. Daises cost $2 a piece and each rose costs $3 a piece. Martha doesn't want to spend over $500 on her garden
Mathematics
1 answer:
valentinak56 [21]2 years ago
5 0

Answer:

150 daisies and 66 roses.

Step-by-step explanation:

To carry out the exercise, you have to propose equations with the requirements of Martha, a garden of 400 square feet and not spend more than $ 500. Let X be the number of daisies and Y the number of roses. We have left that:

2 * X + 1.5 * Y = 400 (1)

2 * X + 3 * Y = 500 (2)

We have two equations with two unknowns, therefore we proceed to solve. If we subtract (1) in (2) we have:

2 * X + 3 * Y - 2 * X - 1.5 * Y = 500 - 400, rearranging we have:

1.5 * Y = 100

Y = 100 / 1.5 = 66.7, it would be approximately 66 roses.

Replacing the value of Y now in (1) we have:

2 * X + 1.5 * 66.7 = 400

2 * X + 99.9 = 400

X = (400-99.9) / 2 = 150.05, would be approximately 150 daisies.

We replace in (1) and (2) to check:

2 * 150 + 1.5 * 66 = 399, is the maximum we can spend because one more flower would be spent.

2 * 150 + 3 * 66 = 498, is the maximum we can spend because one more flower would be spent.

Therefore it meets Martha's requirements for his garden, 150 daisies and 66 roses.

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the expected value for the 3 point shot = (3 * 0.30) + (0 * 0.70) = 0.90

the expected value for the 2 point shot = (2 * 0.48) + (0 * 0.52) = 0.96

the expected value for the 2 point shot is higher than the 3 point shot so he should pass the ball

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2 years ago
Three couples and two single individuals have been invited to an investment seminar and have agreed to attend. Suppose the proba
Maksim231197 [3]

Answer:

(a) Probability mass function

P(X=0) = 0.0602

P(X=1) = 0.0908

P(X=2) = 0.1704

P(X=3) = 0.2055

P(X=4) = 0.1285

P(X=5) = 0.1550

P(X=6) = 0.1427

P(X=7) = 0.0390

P(X=8) = 0.0147

NOTE: the sum of the probabilities gives 1.0068 for rounding errors. It can be divided by 1.0068 to get the adjusted values.

(b) Cumulative distribution function of X

F(X=0) = 0.0602

F(X=1) = 0.1510

F(X=2) = 0.3214

F(X=3) = 0.5269

F(X=4) = 0.6554

F(X=5) = 0.8104

F(X=6) = 0.9531

F(X=7) = 0.9921

F(X=8) = 1.0068

Step-by-step explanation:

Let X be the number of people who arrive late to the seminar, we can assess that X can take values from 0 (everybody on time) to 8 (everybody late).

<u>For X=0</u>

This happens when every couple and the singles are on time (ot).

P(X=0)=P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot)\\\\P(X=0)=(1-0.43)^{5}=0.57^5= 0.0602

<u>For X=1</u>

This happens when only one single arrives late. It can be #4 or #5. As the probabilities are the same (P(#4=late)=P(#5=late)), we can multiply by 2 the former probability:

P(X=1) = P(\#4=late)+P(\#5=late)=2*P(\#4=late)\\\\P(X=1) = 2*P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=late)*P(\#5=ot)\\\\P(X=1) = 2*0.57*0.57*0.57*0.43*0.57\\\\P(X=1) = 2*0.57^4*0.43=2*0.0454=0.0908

<u>For X=2</u>

This happens when

1) Only one of the three couples is late, and the others cooples and singles are on time.

2) When both singles are late , and the couples are on time.

P(X=2)=3*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=ot)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=2)=3*(0.43*0.57^4)+(0.43^2*0.57^3)=0.1362+0.0342=0.1704

<u>For X=3</u>

This happens when

1) Only one couple (3 posibilities) and one single are late (2 posibilities). This means there are 3*2=6 combinations of this.

P(X=3)=6*(P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=3)=6*(0.43^2*0.57^3)=6*0.342=0.2055

<u>For X=4</u>

This happens when

1) Only two couples are late. There are 3 combinations of these.

2) Only one couple and both singles are late. Only one combination of these situation.

P(X=4)=3*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=ot)*P(\#5=ot))+P(\#1=l)*P(\#2=ot)*P(\#3=ot)*P(\#4=l)*P(\#5=l)\\\\P(X=4)=3*(0.43^2*0.57^3)+(0.43^3*0.57^2)\\\\P(X=4)=3*0.0342+ 0.0258=0.1027+0.0258=0.1285

<u>For X=5</u>

This happens when

1) Only two couples (3 combinations) and one single are late (2 combinations). There are 6 combinations.

P(X=6)=6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=6*(0.43^3*0.57^2)=6*0.0258=0.1550

<u>For X=6</u>

This happens when

1) Only the three couples are late (1 combination)

2) Only two couples (3 combinations) and one single (2 combinations) are late

P(X=6)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=ot)*P(\#5=ot)+6*(P(\#1=l)*P(\#2=l)*P(\#3=ot)*P(\#4=l)*P(\#5=ot))\\\\P(X=6)=(0.43^3*0.57^2)+6*(0.43^4*0.57)\\\\P(X=6)=0.0258+6*0.0195=0.0258+0.1169=0.1427

<u>For X=7</u>

This happens when

1) Only one of the singles is on time (2 combinations)

P(X=7)=2*P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=ot)\\\\P(X=7)=2*0.43^4*0.57=0.0390

<u>For X=8</u>

This happens when everybody is late

P(X=8)=P(\#1=l)*P(\#2=l)*P(\#3=l)*P(\#4=l)*P(\#5=l)\\\\P(X=8) = 0.43^5=0.0147

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2 years ago
A school principal has to choose randomly among the six best students in each grade to be the school captain every month. For th
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Since there are 6 students out of which one needs to be selected, the principal chose two die on which there are six numbers each numbered from 1 , 2, 3, 4, 5, 6.

Since there are two dice, the total possible outcome is 36.

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Hence, the principal used a fair method because each result is an equally likely possible outcome.

Option B is correct.

5 0
2 years ago
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babymother [125]

Answer:

50 years

Step-by-step explanation:

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2 years ago
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RSB [31]

Answer:

Pr(X-Y ≤ 44.2) = 0.5593

Step-by-step explanation:

for a certain breed of terrier

Mean(μ) = 72cm

Standard deviation (σ) = 10cm

n = 64

For a certain breed of poodle

Mean(μ) = 28cm

Standard deviation (σ) = 5cm

n = 100

Let X be the random variable for the height of a certain breed of terrier

Let Y be the random variable for the height of a certain breed of poodle

μx - μy = 72 -28

= 44

σx - σy = √(σx^2/nx + σy^2/ny)

= √10^2/64 + 5^2/100

= √100/64 + 25/100

= √ 1.8125

= 1.346

Using normal distribution,

Z= (X-Y- μx-y) / σx-y

Z= (44.2 - 44) / 1.346

Z= 0.2/1.346

Z= 0.1486

From the Z table, Z = 0.149 = 0.0593

Φ(z) = 0 0593

The probability that the difference of the observed sample mean is at most 44.3 is Pr(Z ≤ 44.2)

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Pr(Z≤a) = 0.5 + Φ(z)

Pr(Z ≤ 44.2) = 0.5 + 0.0593

= 0.5593

Therefore,

Pr(X-Y ≤ 44.2) = 0.5593

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