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3 years ago

. Write the following expression in factored form using the greatest common factor. 50a3 + 10a2

1 answer:

3 0

The factors of 50a³ are 1, 2, 5, 10, 25, 50,
   and their products with a, a² and a³ .

The factors of 10a² are 1, 2, 5, 10,
    and their products with 'a' and a² .

Their common factors are 1, 2, 5, 10,
   and their products with 'a' and a².

Their greatest common factor is 10a² .

(Another way to spot it, easily, is to remember this helpful factoid:

   If the smaller number is a factor of the larger number,
   then the smaller number is their greatest common factor.

Using the greatest common factor, then . . .

   50a³ + 10a² =   10a²(5a + 1) .

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Ebony earned scores of 72 70, 81, 99, and 88 on five math tests. Determine her

Minchanka [31]

Let it be x

$\boxed{\sf Average=\dfrac{Sum\:of\:terms}{No\:of\:terms}}$

$\\ \sf\longmapsto \dfrac{72+70+81+99+88+x}{6}=83$

$\\ \sf\longmapsto \dfrac{410+x}{6}=83$

$\\ \sf\longmapsto 410+x=498$

$\\ \sf\longmapsto x=498-410$

$\\ \sf\longmapsto x=88$

6 0

3 years ago

before sales tax,the classical music CD cost $15.Jan gave the clerk a $20 bill and recieved $4.17 change.what is the rate of sal

Harman [31]

B.4% at least that's what I got lol I did it at least twice in my head

6 0

3 years ago

Add [1/-4 3/5] [-2/6 -2/4]

brilliants [131]

Answer:

25/138

Step-by-step explanation:

1 Convert 4\frac{3}{5}4

to improper fraction. Use this rule: a \frac{b}{c}=\frac{ac+b}{c}a

ac+b

\frac{1}{-(\frac{4\times 5+3}{5})}(-\frac{2}{6}-\frac{2}{4})

−(

4×5+3

)

(−

−

)

2 Simplify 4\times 54×5 to 2020.

\frac{1}{-(\frac{20+3}{5})}(-\frac{2}{6}-\frac{2}{4})

−(

20+3

)

(−

−

)

3 Simplify 20+320+3 to 2323.

\frac{1}{-(\frac{23}{5})}(-\frac{2}{6}-\frac{2}{4})

−(

)

(−

−

)

4 Simplify \frac{2}{6}

to \frac{1}{3}

\frac{1}{-(\frac{23}{5})}(-\frac{1}{3}-\frac{2}{4})

−(

)

(−

−

)

5 Simplify \frac{2}{4}

to \frac{1}{2}

\frac{1}{-(\frac{23}{5})}(-\frac{1}{3}-\frac{1}{2})

−(

)

(−

−

)

6 Find the Least Common Denominator (LCD) of \frac{1}{3},\frac{1}{2}

. In other words, find the Least Common Multiple (LCM) of 3,23,2.

LCD = 66

7 Make the denominators the same as the LCD.

-\frac{1\times 2}{3\times 2}-\frac{1\times 3}{2\times 3}−

3×2

1×2

−

2×3

1×3

8 Simplify. Denominators are now the same.

-\frac{2}{6}-\frac{3}{6}−

−

9 Join the denominators.

\frac{-2-3}{6}

−2−3

10 Simplify -\frac{1}{3}-\frac{1}{2}−

−

to -\frac{5}{6}−

\frac{1}{-(\frac{23}{5})}\times \frac{-5}{6}

−(

)

−5

11 Move the negative sign to the left.

-\frac{1}{\frac{23}{5}}\times \frac{-5}{6}−

−5

12 Invert and multiply.

-\frac{5}{23}\times \frac{-5}{6}−

−5

13 Use this rule: \frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}

-\frac{5\times -5}{23\times 6}−

23×6

5×−5

14 Simplify 5\times -55×−5 to -25−25.

-\frac{-25}{23\times 6}−

23×6

−25

15 Simplify 23\times 623×6 to 138138.

-\frac{-25}{138}−

138

−25

16 Move the negative sign to the left.

-(-\frac{25}{138})−(−

138

)

17 Remove parentheses.

\frac{25}{138}

138

7 0

3 years ago

2(2t+4)=3/4(24−8t) please help

Tcecarenko [31]

Step-by-step explanation:

the process is shown above hope you understand.

5 0

3 years ago

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

7 0

2 years ago