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2 years ago

Solve 4a = 20

Mathematics

2 answers:

Deffense [45]2 years ago

7 0

Answer:

1)4a=20

solution:a=20-4

a=16

2)3y+9=24

solution:3y=24-9

3y=15

y=15/3

y= 5

3)y/3-5=4

solution:y/3=4+5

y/3=9

y=9×3

y=27

4)3=9-4k

solution:3=5k

3-5=k

-2=k

k=-2

5)3x+8=2

solution:3x=2-8

3x=-5

x=-5/3

x= -1.67

Step-by-step explanation:

plz Mark my answer in brainlist

olya-2409 [2.1K]2 years ago

6 0

Here's the solutions :

1. solve for a :

$4a = 20$

$a = \dfrac{20}{4}$

$a = 5$

2. solve for y :

$3y + 9 = 24$

$3y = 24 - 9$

$3y = 15$

$y = \dfrac{15}{3}$

$y = 5$

3. solve for y :

$\dfrac{y}{3} - 5 = 4$

$\dfrac{y}{3} = 5 + 4$

$\dfrac{y}{3} = 9$

$y = 9 \times 3$

$y = 27$

4. solve for k :

$3 = 9 - 4k$

$3 - 9 = -4 k$

$- 6 = - 4 k$

$k = \dfrac{ - 6}{ - 4}$

$k = \dfrac{3}{2} \: \: \: or \: \: 1.5$

5. there's something wrong with this question.

6. solve for x :

$3x + 8 = 2$

$3x = 2 - 8$

$3x = - 6$

$x = \dfrac{ - 6}{3}$

$x = - 2$

i hope it helped...

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Regular triangular pyramid has 6 cm long base edge and slant height k=9 cm. Find the lateral area of the pyramid.

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Answer:

81 cm²

Step-by-step explanation:

Since, the lateral face of a triangular pyramid is a triangle,

Given,

The base edge or the base of one lateral face of pyramid, a = 6 cm,

And, the slant height or the height of the face, k = 9 cm,

Thus, the area of one lateral face of the pyramid,

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3 years ago

Read 2 more answers

The taxi and takeoff time for commercial jets is a random variable x with a mean of 8.9 minutes and a standard deviation of 2.9

Eva8 [605]

Answer:

a) 0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

b) 0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes

c) 0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

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.

Central Limit Theorem

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mean of 8.9 minutes and a standard deviation of 2.9 minutes.

This means that $\mu = 8.9, \sigma = 2.9$

Sample of 37:

This means that $n = 37, s = \frac{2.9}{\sqrt{37}}$

(a) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes?

320/37 = 8.64865

Sample mean below 8.64865, which is the p-value of Z when X = 8.64865. So

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{8.64865 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -0.53$

$Z = -0.53$ has a p-value of 0.2981

0.2981 = 29.81% probability that for 37 jets on a given runway, total taxi and takeoff time will be less than 320 minutes.

(b) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes?

275/37 = 7.4324

Sample mean above 7.4324, which is 1 subtracted by the p-value of Z when X = 7.4324. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{7.4324 - 8.9}{\frac{2.9}{\sqrt{37}}}$

$Z = -3.08$

$Z = -3.08$ has a p-value of 0.001

1 - 0.001 = 0.999

0.999 = 99.9% probability that for 37 jets on a given runway, total taxi and takeoff time will be more than 275 minutes.

(c) What is the probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes?

Sample mean between 7.4324 minutes and 8.64865 minutes, which is the p-value of Z when X = 8.64865 subtracted by the p-value of Z when X = 7.4324. So

0.2981 - 0.0010 = 0.2971

0.2971 = 29.71% probability that for 37 jets on a given runway, total taxi and takeoff time will be between 275 and 320 minutes

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