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natta225 [31]
2 years ago
10

andrew rides his bike at a speed of 28 km per hour. he rides for one amd one quarter hours at this speed without stopping. How f

ar does he ride?
Mathematics
2 answers:
quester [9]2 years ago
8 0

Answer:

Well.

Step-by-step explanation:

If its 28 km for hour. A quarter of 28 is 7 so then 28+7=35 km.

RoseWind [281]2 years ago
6 0

Answer: 35 Km

Step-by-step explanation:

28 Km in 60 minutos. How many Km. In 75 minutes (One and one quarter hour =75min)

28/x = 60/75 (cross multiplication)

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Which of these numbers is GREATEST? A) 5.7 × 106 B) 5.7 × 107 C) 5.07 × 106 D) 5.07 × 107
Vaselesa [24]

Answer:

B) 5.7 × 107 = 609.9

Step-by-step explanation:

A) 604.2

B) 609.9

C) 537.42

D) 542.49

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3 years ago
If a circle has a diameter of 36cm, what is its exact area?
r-ruslan [8.4K]

Answer: 1017.88cm²

Step-by-step explanation:

5 0
2 years ago
Read 2 more answers
3(x+3)+3x=15<br><br> Solution and check<br><br> Shown in attachment below
OLga [1]
3x + 9 + 3x =15
6x + 9 = 15
- 9 = -9
6x = 6
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6 0
3 years ago
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
The circumference of a circle is 12pi feet. What is the diameter?
zaharov [31]

Answer:The diameter is 12cm

Step-by-step explanation:

4 0
3 years ago
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