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Fittoniya [83]
3 years ago
12

What is 2/7 plus 8/21

Mathematics
1 answer:
Alchen [17]3 years ago
7 0

Answer:

2/3

Step-by-step explanation

21 is a multiple of 7. To add fractions you must have the same denominator. 7 times 3 is 21. So multiple the whole fraction, both numerator and denominator(2 and 7) by 3 to get 6/21. From there just add 6/21 and 8/21 together to get 14/21 then in they have a GCF(greatest common factor), which in this case is 7. Divide it by the GCF to simplify the fraction.

Hope this helps!

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ramp with a 24 degree angle of elevation from the ground. How tall will the ramp rise from the ground at its tallest end?
Step2247 [10]

Answer:

12

Step-by-step explanation:

5 0
3 years ago
Find the missing side
nevsk [136]

Answer:

9cm

Step-by-step explanation:

Area of trapezium=(1/2)×(a+b)h

A=33

a=2

b=?

h=6

(1/2)(2+b)(6)=33

(1/2)(12+6b)=33

6+3b=33

3b=33-6

3b=27

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3 0
3 years ago
What is the slope of the line through (-1,4) and (5,2)
butalik [34]

Answer:

Slope (m) = -\frac{1}{3}

Step-by-step explanation:

Use the following equation:

Slope (m) = \frac{y_2 - y_1}{x_2 - x_1}

Let:

(x_1 , y_1) = (-1 , 4)\\(x_2 , y_2) = (5 , 2)

Plug in the corresponding numbers to the corresponding variables:

m = \frac{2 - 4}{5 - (-1)} = \frac{2 - 4}{5 + 1} = \frac{-2}{6}

Simplify. Divide common factors from both the numerator and denominator:

\frac{(-\frac{2}{6})}{\frac{2}{2} } = -\frac{1}{3}

-\frac{1}{3} is your answer.

~

8 0
3 years ago
A shop sells a particular of video recorder. Assuming that the weekly demand for the video recorder is a Poisson variable with t
julia-pushkina [17]

Answer:

a) 0.5768 = 57.68% probability that the shop sells at least 3 in a week.

b) 0.988 = 98.8% probability that the shop sells at most 7 in a week.

c) 0.0104 = 1.04% probability that the shop sells more than 20 in a month.

Step-by-step explanation:

For questions a and b, the Poisson distribution is used, while for question c, the normal approximation is used.

Poisson distribution:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

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

In which

x is the number of successes

e = 2.71828 is the Euler number

\lambda is the mean in the given interval.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

The Poisson distribution can be approximated to the normal with \mu = \lambda, \sigma = \sqrt{\lambda}, if \lambda>10.

Poisson variable with the mean 3

This means that \lambda= 3.

(a) At least 3 in a week.

This is P(X \geq 3). So

P(X \geq 3) = 1 - P(X < 3)

In which:

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

Then

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

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

P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240

So

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0498 + 0.1494 + 0.2240 = 0.4232

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 1 - 0.4232 = 0.5768

0.5768 = 57.68% probability that the shop sells at least 3 in a week.

(b) At most 7 in a week.

This is:

P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)

In which

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

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

P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240

P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240

P(X = 4) = \frac{e^{-3}*3^{4}}{(4)!} = 0.1680

P(X = 5) = \frac{e^{-3}*3^{5}}{(5)!} = 0.1008

P(X = 6) = \frac{e^{-3}*3^{6}}{(6)!} = 0.0504

P(X = 7) = \frac{e^{-3}*3^{7}}{(7)!} = 0.0216

Then

P(X \leq 7) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7) = 0.0498 + 0.1494 + 0.2240 + 0.2240 + 0.1680 + 0.1008 + 0.0504 + 0.0216 = 0.988

0.988 = 98.8% probability that the shop sells at most 7 in a week.

(c) More than 20 in a month (4 weeks).

4 weeks, so:

\mu = \lambda = 4(3) = 12

\sigma = \sqrt{\lambda} = \sqrt{12}

The probability, using continuity correction, is P(X > 20 + 0.5) = P(X > 20.5), which is 1 subtracted by the p-value of Z when X = 20.5.

Z = \frac{X - \mu}{\sigma}

Z = \frac{20 - 12}{\sqrt{12}}

Z = 2.31

Z = 2.31 has a p-value of 0.9896.

1 - 0.9896 = 0.0104

0.0104 = 1.04% probability that the shop sells more than 20 in a month.

5 0
3 years ago
Sydney rolls a ball 63 inches across the floor. This is 7 times the distance that Brendan rolls his ball across the floor. Let n
Travka [436]

9 inches across the floor Brendan rolls his ball if Sydney rolls a ball 63 inches across the floor.

<h3>What is linear equation?</h3>

It is defined as the relation between two variables, if we plot the graph of the linear equation we will get a straight line.

If in the linear equation, one variable is present, then the equation is known as the linear equation in one variable.

We have:

Sydney rolls a ball 63 inches across the floor. This is 7 times the distance that Brendan rolls his ball across the floor.

Sydney rolls = 7n

Here n represent the number of inches Brendan rolls his ball.

63 = 7n

n = 63/7 = 9 inches

Thus, 9 inches across the floor Brendan rolls his ball if Sydney rolls a ball 63 inches across the floor.

Learn more about the linear equation here:

brainly.com/question/11897796

#SPJ1

8 0
2 years ago
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