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adelina 88 [10]
3 years ago
11

Factor each completely. 10a² - 9a + 2 ​

Mathematics
1 answer:
Sophie [7]3 years ago
3 0

Answer:

a= \frac{2}{5} or \frac{1}{2}

Step-by-step explanation:

Equate the equation to zero

10a^{2}-9a+2=0

10a^{2}-5a-4a+2=0

5a(2a-1)-2(2a-1)=0

Write out the factors

(5a-2)(2a-1)=0

Then equate each factor to zero

5a-2=0 OR 2a-1=0

5a=2  OR  2a=1

a= \frac{2}{5}   OR a= \frac{1}{2}

a=\frac{2}{5} or \frac{1}{2}

You might be interested in
The International Air Transport Association surveys business travelers to develop quality ratings for transatlantic gateway airp
motikmotik

Answer:

The 95% confidence interval for the population mean rating is (5.73, 6.95).

Step-by-step explanation:

We start by calculating the mean and standard deviation of the sample:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{50}(6+4+6+. . .+6)\\\\\\M=\dfrac{317}{50}\\\\\\M=6.34\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{49}((6-6.34)^2+(4-6.34)^2+(6-6.34)^2+. . . +(6-6.34)^2)}\\\\\\s=\sqrt{\dfrac{229.22}{49}}\\\\\\s=\sqrt{4.68}=2.16\\\\\\

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=6.34.

The sample size is N=50.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

s_M=\dfrac{s}{\sqrt{N}}=\dfrac{2.16}{\sqrt{50}}=\dfrac{2.16}{7.071}=0.305

The degrees of freedom for this sample size are:

df=n-1=50-1=49

The t-value for a 95% confidence interval and 49 degrees of freedom is t=2.01.

The margin of error (MOE) can be calculated as:

MOE=t\cdot s_M=2.01 \cdot 0.305=0.61

Then, the lower and upper bounds of the confidence interval are:

LL=M-t \cdot s_M = 6.34-0.61=5.73\\\\UL=M+t \cdot s_M = 6.34+0.61=6.95

The 95% confidence interval for the mean is (5.73, 6.95).

4 0
3 years ago
(2, -4), (x, y) , A , and B all lie on the line. Find an equation relating x and y.​
frutty [35]

Answer:

An equation relating x and y is:

y = 3/4x - 5.5

Step-by-step explanation:

The slope-intercept form of the line equation

y = mx+b

where

  • m is the slope
  • b is the y-intercept

Given the points

(2, -4), (x, y), A(6, -1), and B(10, 2)

Taking any two points let say A(6, -1), and B(10, 2) to determine the slope

\mathrm{Slope}=\frac{y_2-y_1}{x_2-x_1}

\left(x_1,\:y_1\right)=\left(6,\:-1\right),\:\left(x_2,\:y_2\right)=\left(10,\:2\right)

m=\frac{2-\left(-1\right)}{10-6}

Refine

m=\frac{3}{4}

We know that the value of the y-intercept can be determined by setting x = 0, and determining the corresponding value of y.

From the graph, it is clear

at x = 0, y = -5.5

Thus, the y-intercept b = -5.5

Thus, substituting b = -5.5 and m = 3/4 in the slope-intercept form of the line equation

y = mx+b

y = 3/4x + (-5.5)

y = 3/4x - 5.5

Therefore, an equation relating x and y is:

y = 3/4x - 5.5

8 0
3 years ago
How to find area of 4-sided figure
hoa [83]

Answer:

Step-by-step explanation:

U do bass times height to get ure answer then if it is a square multiply it one time by itself. And if it is a triangle multiply bass times height divided by two and u will get ure answer

8 0
3 years ago
Whats the common factor of 18,27
Leviafan [203]

Well, it depends, there may be multiple common factors, if it's the greatest common factor, then there is only one.  The way to do this is to list all the factors of each number.


So Factors of

18: 1,2,3,6,9,18

27: 1,3,9,27

Common Factors : 1, 3, 9

Greatest Common Factor: 9

5 0
3 years ago
John, Joe, and James go fishing. At the end of the day, John comes to collect his third of the fish. However, there is one too m
Dmitry [639]

Answer:

The minimum possible initial amount of fish:52

Step-by-step explanation:

Let's start by saying that

x = is the initial number of fishes

John:

When John arrives:

  • he throws away one fish from the bunch

x-1

  • divides the remaining fish into three.

\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}

  • takes a third for himself.

\dfrac{x-1}{3} + \dfrac{x-1}{3}

the remaining fish are expressed by the above expression. Let's call it John

\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}

and simplify it!

\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}

When Joe arrives:

  • he throws away one fish from the remaining bunch

\text{John} -1

  • divides the remaining fish into three

\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}

  • takes a third for himself.

\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

the remaining fish are expressed by the above expression. Let's call it Joe

\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}

and simiplify it

\text{Joe}=\dfrac{2}{3}(\text{John}-1)

since we've already expressed John in terms of x, we express the above expression in terms of x as well.

\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)

\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}

When James arrives:

We're gonna do this one quickly, since its the same process all over again

\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}

\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)

\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}

This is the last remaining pile of fish.

We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.

Whatever the last remaining pile was (let's say n), a third is taken away by James. the remaining bunch would be \frac{n}{3}+\frac{n}{3}

hence we've expressed the last pile in terms of n as well.  Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:

\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}

L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..

at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)

14=\dfrac{8x}{27} - \dfrac{38}{27}

by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!

14=\dfrac{8x}{27} - \dfrac{38}{27}

x=52

This is minimum possible amount of fish before John threw out the first fish

8 0
3 years ago
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