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Pani-rosa [81]
2 years ago
7

Bouquets are to be made using white tulips and red tulips, and the ratio of the number of white tulips to the number of red tuli

ps is to be the same in each bouquet. If there are 15 white tulips and 85 red tulips available for the bouquets, what is the greatest number of bouquets that can be made using all the tulips available?A. 3B. 5C. 8D. 10E. 13
Mathematics
1 answer:
zlopas [31]2 years ago
4 0

Answer:

B. 5

Step-by-step explanation:

The highest common factor between 85 and 15 = 5

therefore the ratio of red to white tulip = 17:3

That is red tulip ratio 85/5 = 17

White tulip ratio 15/5 = 3

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(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX
pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(c) 0.3707

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and  variance <em>σ</em>² = 36.

(a)

Compute the value of P (X > 5) as follows:

P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

P(4

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

P(X

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

P(X

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z

Thus, the value of P (X > 16) is 0.1587.

**Use a <em>z</em>-table for the probabilities.

8 0
2 years ago
A semicircular flower bed has a diameter of 3 metres. What is the area of the flower bed?
Yuri [45]

The area of circle is

A_{\text{circle}}=\pi r^2.

Then the area of semicircular region is

A_{\text{semicircular region}}=\dfrac{1}{2} \pi r^2.

If r=3 m, you have

A_{\text{semicircular region}}=\dfrac{1}{2} \pi \cdot 3^2=\dfrac{9\pi}{2}=4.5\pi \approx 14.13 sq. m.

Answer: A_{\text{semicircular region}}=4.5\pi \approx 14.13 sq. m.

3 0
2 years ago
Read 2 more answers
Mrs.Gunn makes one costume in 3hours and 20 minutes. How long will it take her to make 6 costumes for a school play?
qaws [65]

Answer:

20 hours

Step-by-step explanation:

4 0
2 years ago
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Folosind procedeul mersului invers afla pentru can numarul stiind ca: a) daca la acel numar aduni 1345 apoi scazi 155 obtii numa
earnstyle [38]
Yeah I think so you have to solve and break it down
7 0
2 years ago
According to a study in a medical journal, 202 of a sample of 5,990 middle-aged men had developed diabetes. It also found that m
tekilochka [14]

Answer:

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Has diabetes.

Event B: Is very active.

Probability of having diabetes:

To find this probability, we take in consideration that:

It also found that men who were very active (burning about 3,500 calories daily) were a fourth as likely to develop diabetes compared with men who were sedentary. Assume that one-fifth of all middle-aged men are very active, and the rest are classified as sedentary.

So the probability of developing diabetes is:

x of 4/5 = x of 0.8(not active)

x/4 = 0.25x of 1/5 = 0.2(very active). So

P(A) = 0.8x + 0.25*0.2x = 0.85x

Probability of developing diabetes while being very active:

0.25x of 0.2. So

P(A \cap B) = 0.25x*0.2 = 0.05x

What is the probability that a middle-aged man with diabetes is very active?

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.05x}{0.85x} = \frac{0.05}{0.85} = 0.0588

0.0588 = 5.88% probability that a middle-aged man with diabetes is very active

4 0
2 years ago
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