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Gelneren [198K]
3 years ago
15

Nick found the quotient of 8.64 and 1.25 × 105. His work is shown below. 1. (8.64 × 101) (1.25 × 105) 2. (8.64 1.25 ) (101 105 )

3. 6.912 × 10–4 Analyze Nick’s solution. Is he correct? If not, what was his mistake? Yes, Nick is correct. No, the power multiplied to 8.64 should have an exponent of 0. No, he needed to subtract the coefficients. No, he needed to add the exponents.
Mathematics
2 answers:
docker41 [41]3 years ago
8 0

Answer:

B.) No, the power multiplied to 8.64 should have an exponent of 0.

Step-by-step explanation:

I just took the assignment on edgenuity!!

Savatey [412]3 years ago
4 0

Answer:

B.) No, the power multiplied to 8.64 should have an exponent of 0.

Step-by-step explanation:

took the test on Edgenuity and got it right! Hope this helps and please mark brainliest if can.

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Nicholas wrote the steps below to simplify the fraction 20/30 find his error and correct it
Dmitrij [34]
Unfortunately, Nicholas' steps in simplifying the fraction is not reflected in the question. However, the general idea of the simplifying fraction is to find the greatest common factor (GCF) of both the numerator and the denominator. In this case, the GCF is 10. Then, divide both numbers by the GCF. The answer for this item should be 2/3. 
7 0
3 years ago
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A farmer has a collapsable grain bag which can expand both vertically and horizontally as he needs it to. He wants to find a fun
bija089 [108]

Answer:

3x^{3}+15x^{2}+x+5

Step-by-step explanation:

We are told that the area of the bag can be represented by the function f(x)=3x^{2} +1 and as the bag is raised up its height can be represented by the function g(x)=x +5.

Since we know that we can find volume of cuboid by multiplying base area to its height. We are given area and height of bag as functions. Now we will find volume of the bag by multiplying these functions.  

\text{Volume of collapsible bag}=f(x)*g(x)  

\text{Volume of collapsible bag}=(3x^{2}+1)*(x +5)

After using distributive property we will get,

\text{Volume of collapsible bag}=3x^{2}(x+5)+1(x+5)

\text{Volume of collapsible bag}=3x^{3}+15x^{2}+x+5

Therefore, the collapsible bag can hold 3x^{3}+15x^{2}+x+5 grain.

6 0
3 years ago
Is will like to make this as simple as i can. -1 x 372 - -31= what
mario62 [17]
The answer is -403

The order of operations puts multiplication before subtraction. -1 x 372= -372, and -372-31= -403
5 0
3 years ago
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One card is drawn from a well shuffled deck of 52 cards find the probability of getting a red 10
Julli [10]
The probability of getting a red card is 1/2 and the probability of getting a 10 is 4/52. To find the probability of getting both a red card and a 10, multiply the two.

(1/2)(4/52) = 4/104
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Therefore, the probability is 1/26
7 0
3 years ago
Suppose small aircraft arrive at a certain airport according to a Poisson process with rate a 5 8 per hour, so that the number o
timurjin [86]

Answer:

(a) P (X = 6) = 0.12214, P (X ≥ 6) = 0.8088, P (X ≥ 10) = 0.2834.

(b) The expected value of the number of small aircraft that arrive during a 90-min period is 12 and standard deviation is 3.464.

(c) P (X ≥ 20) = 0.5298 and P (X ≤ 10) = 0.0108.

Step-by-step explanation:

Let the random variable <em>X</em> = number of aircraft arrive at a certain airport during 1-hour period.

The arrival rate is, <em>λ</em>t = 8 per hour.

(a)

For <em>t</em> = 1 the average number of aircraft arrival is:

\lambda t=8\times 1=8

The probability distribution of a Poisson distribution is:

P(X=x)=\frac{e^{-8}(8)^{x}}{x!}

Compute the value of P (X = 6) as follows:

P(X=6)=\frac{e^{-8}(8)^{6}}{6!}\\=\frac{0.00034\times262144}{720}\\ =0.12214

Thus, the probability that exactly 6 small aircraft arrive during a 1-hour period is 0.12214.

Compute the value of P (X ≥ 6) as follows:

P(X\geq 6)=1-P(X

Thus, the probability that at least 6 small aircraft arrive during a 1-hour period is 0.8088.

Compute the value of P (X ≥ 10) as follows:

P(X\geq 10)=1-P(X

Thus, the probability that at least 10 small aircraft arrive during a 1-hour period is 0.2834.

(b)

For <em>t</em> = 90 minutes = 1.5 hour, the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 1.5=12

The expected value of the number of small aircraft that arrive during a 90-min period is 12.

The standard deviation is:

SD=\sqrt{\lambda t}=\sqrt{12}=3.464

The standard deviation of the number of small aircraft that arrive during a 90-min period is 3.464.

(c)

For <em>t</em> = 2.5 the value of <em>λ</em>, the average number of aircraft arrival is:

\lambda t=8\times 2.5=20

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

P(X\geq 20)=1-P(X

Thus, the probability that at least 20 small aircraft arrive during a 2.5-hour period is 0.5298.

Compute the value of P (X ≤ 10) as follows:

P(X\leq 10)=\sum\limits^{10}_{x=0}(\frac{e^{-20}(20)^{x}}{x!})\\=0.01081\\\approx0.0108

Thus, the probability that at most 10 small aircraft arrive during a 2.5-hour period is 0.0108.

8 0
3 years ago
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