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Margarita [4]
3 years ago
10

Container X has 2 cups of water with 6 teaspoons of sugar. Container Y has 4 cups of water with 8 teaspoons of sugar. In which c

ontainer is the water sweeter? A. Container x B.Container Y C. Both containers have water that is equally sweet. D. (Cannot say without tasting the water in each container).
Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
6 0

Answer:

A) Container X

Step-by-step explanation:

X=2C:6S  ratio, now reduce amount but keep the same ratio

X=1C:3S  

Y=4C:8S  ratio, now reduce amount but keep the same ratio

Y=1C:2S

Container X has the sweeter water because it's ratio of sugar to water is higher.

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Simplify.<br> -5x^4(-3x^2+4x-2)
dybincka [34]
Simplification/ 15x^6-20x^5+10x^4

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
Please help don’t get it ASAP
maxonik [38]

Answer:

Option C is correct.

38 miles

Step-by-step explanation:

Let x be the distance cover by Michelle

Given:

A circle diameter is 3 miles.

And Michelle biked around the trail 4 times.

We know that

The circumference of a circle.

circumference = 2\pi r----------(1)

where r=\frac{Diameter}{2}

r=\frac{3}{2} \\r=1.5 miles

Put \pi =3.14 and r value in equation 1.

circumference = 2\times 3.14 1.5

circumference = 9.42 mil;es

The circumference of a circle is 9.42 miles.

And Michelle biked 4 times, that is equal to 4 \times circumference.

x = 4\times 9.42

x = 37.65 miles.

That is equal to 38 miles.

Therefore, Michelle biked around 4 trail is 38 miles.

5 0
3 years ago
A bee flies 20 feet per second directly to a flowerbed from its hive. The bee stays at the flowerbed for fifteen minutes. Then i
Ksenya-84 [330]

A. The equation to find the distance of the flowerbed from the hive is: \frac{x}{20}+\frac{x}{12}+900= 1200 , where x is the distance.

B.  The flowerbed is 2250 feet far from the hive.

<u><em>Explanation</em></u>

Suppose, the distance of the flowerbed from the hive is x feet.

The bee flies 20 feet per second directly to a flowerbed from its hive and flies directly back to the hive at 12 feet per second.

As, Time=\frac{Distance}{Speed}

So, the time taken by the bee to reach the flowerbed = \frac{x}{20} seconds and the time taken to fly back to the hive =\frac{x}{12} seconds.

The bee stays at the flowerbed for 15 minutes or (15×60)seconds or 900 seconds and it is away from the hive for a total of 20 minutes or (20×60)seconds or 1200 seconds.

So, the equation will be.......

\frac{x}{20}+\frac{x}{12}+900= 1200\\ \\ \frac{3x+5x}{60}= 300\\ \\ 8x= 60*300\\ \\ 8x=18000\\ \\ x=\frac{18000}{8}=2250

Thus, the distance of the flowerbed from the hive is 2250 feet.

5 0
4 years ago
The Wongs are putting a flagstone deck around their pool. The pool is rectangular. Its dimensions are 8 m by 4 m. The deck will
VLD [36.1K]

Answer: 64 i believe

Hope this helps :)

3 0
3 years ago
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