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3 years ago

HELP ASAP!!Solve for x. Enter the solutions from least to greatest.

1 answer:

4 0

X^2 - 5x - 14 = 0

what multiplies to negative 14 and adds to negative 5? -7 and positive 2.

(x - 7) (x + 2) = 0

Lesser x = -2
Greater x = 7

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A farmer has 20 boxes of eggs

aliina [53]

I guess that is 1:10
Hope ur help !;(

5 0

3 years ago

Factor each expression by factoring out the common binomial <br><br> 5a(y + 4) + 8(y + 4)

Tems11 [23]

Answer:

Factoring the term $5a(y + 4) + 8(y + 4)$ we get $(y+4)(5a+8)$

Step-by-step explanation:

We need to factor the term: $5a(y + 4) + 8(y + 4)$

Factoring:

$5a(y + 4) + 8(y + 4)$

Taking (y+4) common

$(y+4)(5a+8)$

It cannot be further factored.

So, Factoring the term $5a(y + 4) + 8(y + 4)$ we get $(y+4)(5a+8)$

6 0

3 years ago

It takes Aaron 5 minutes to squeeze the juice from 3 grapefruits. At this rate, how many minutes will it take him to squeeze the

aliya0001 [1]

20 minutes because 12/3=4 and 4x5=20

3 0

3 years ago

I have a two digit number the sum of the digits is 12 the first digit is 4 Less Than 3 times the second what is my number

nignag [31]

Answer:

Step-by-step explanation:

Let the first digit = x

Let the second digit =y

x + y = 12

x + 4 = 3*y

===========

From the second equation, we learn that x = 3y - 4

Put that into the first equation.

3y - 4 + y = 12 Combine the left side

4y - 4 = 12 Add 4 to both sides

4y = 16 Divide by 4

4y/4 = 16/4

y = 4

=================

x + 4 = 12

x + 4 - 4 = 12-4

x = 8

So the number is 84

3 0

3 years ago

The probability density function of the time you arrive at a terminal (in minutes after 8:00 A.M.) is f(x) = 0.1 exp(−0.1x) for

Blababa [14]

$f_X(x)=\begin{cases}0.1e^{-0.1x}&\text{for }x>0\\0&\text{otherwise}\end{cases}$

a. 9:00 AM is the 60 minute mark:

$f_X(60)=0.1e^{-0.1\cdot60}\approx0.000248$

b. 8:15 and 8:30 AM are the 15 and 30 minute marks, respectively. The probability of arriving at some point between them is

$\displaystyle\int_{15}^{30}f_X(x)\,\mathrm dx\approx0.173$

c. The probability of arriving on any given day before 8:40 AM (the 40 minute mark) is

$\displaystyle\int_0^{40}f_X(x)\,\mathrm dx\approx0.982$

The probability of doing so for at least 2 of 5 days is

$\displaystyle\sum_{n=2}^5\binom5n(0.982)^n(1-0.982)^{5-n}\approx1$

i.e. you're virtually guaranteed to arrive within the first 40 minutes at least twice.

d. Integrate the PDF to obtain the CDF:

$F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x$

Then the desired probability is

$F_X(30)-F_X(15)\approx0.950-0.777=0.173$

7 0

3 years ago