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2 years ago

Please answer below :-)

Mathematics

2 answers:

prohojiy [21]2 years ago

6 0

Answer:

(x+3)(x+2)

Step-by-step explanation:

The numbers 2 and 3 add to 5 and multiply to 6. Therefore when you put them in a factor box, you will end up with $x^2+5x+6$

sertanlavr [38]2 years ago

6 0

Answer : (x+3) (x+2) :

Step-1 : Multiply the coefficient of the first term by the constant 1 • 6 = 6

Step-2 : Find two factors of 6 whose sum equals the coefficient of the middle term, which is 5 .

-6 + -1 = -7
-3 + -2 = -5
-2 + -3 = -5
-1 + -6 = -7
1 + 6 = 7
2 + 3 = 5 That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, 2 and 3
x2 + 2x + 3x + 6

Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x+2)
Add up the last 2 terms, pulling out common factors :
3 • (x+2)
Step-5 : Add up the four terms of step
(x+3) (x+2)

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Jet001 [13]

Answer: A. Suzy made an error going from Step 1 to Step 2.

Step-by-step explanation: Following the order of operations (PEMDAS), we know that exponents are dealt with before multiplication. Suzy used the distributive property of multiplication going from step 1 to step 2, rather than working with the exponent.

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3 years ago

What is the median of the data set:348,349,354,355,357,359,360,367,37,381,386,388,388

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Answer:

Step-by-step explanation:

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3 years ago

Read 2 more answers

If the graph of the function h defined by h(x)=-3x^2+3 is translated vertically upward by 3 units , it becomes the graph of a fu

Slav-nsk [51]

G(x) = h(x) +3 . . . . . h translated upward by 3 units

g(x) = (-3x² +3) +3
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6 0

3 years ago

Which piece wise function is graphed below

AlekseyPX

Answer:

Option A

Step-by-step explanation:

Here is how to approach the problem:

We see that all our restrictions for all four answer choices are relatively the same with a couple of changes here and there.
One way to eliminate choices would be to look at which restrictions don't match the graph.

At x<-5, there is a linear function that does have a -2 slope and will intersect the x axis at -7. The line ends with an open circle, so any answer choice with a linear restriction of x less than or equal to -5 is wrong. This cancels out choices C and D.

Now we have two choices left.

For the quadratic in the middle, the vertex is at (-2,6) and the vertex is a maximum, meaning our graph needs to have a negative sign in front of the highest degree term. In our case, none of our quadratics left are in standard form, and instead are in vertex form.

Vertext form is f(x) = a(x-h)^2 + k.

h being the x-coordinate of the vertex and k being the y-coordinate.

We know that the opposite of h will be the actual x-coordinate of the vertex, so if our vertex is -2, we will see x+2 inside the parenthesis. This leaves option A as the only correct choice.

3 0

3 years ago

The local branch of the Internal Revenue Service spent an average of 21 minutes helping each of 10 people prepare their tax retu

AURORKA [14]

Answer:

The confidence Interval is [- 0.7053 10.4521]

a: The hypotheses are

H0: μ1=μ2 against the claim Ha :μ1≠μ2

b. The critical value for t∝/2 for 17 d.f t > 2.508 and t < -2.111

c. t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

Step-by-step explanation:

When the standard deviations are not the same then the confidence intervals for mean differences are calculated as

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

x1`= 21 x2`= 27

n1= 10 n2= 14

s1= 5.6 s2= 4.3

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

The t∝/2 for 17 d.f = 2.11

Putting the values

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

(21-27) - 2.11√5.6²/10+ 4.3²/14 < u1-u2 <(21-27) +2.11√5.6²/10+4.3²/14

6- 2.11*2.111 < u1-u2 < ( 6 ) +2.11*2.111

6- 4.4521 < u1-u2 < ( 6 ) +5.294

- 1.5479 < u1-u2 < 10.4521

The confidence Interval is [- 0.7053 10.4521]

a: The hypotheses are

H0: μ1=μ2 against the claim Ha :μ1≠μ2

The claim is that there is a difference in the average time spent by the two services

b. The critical value for t∝/2 for 17 d.f t > 2.508 and t < -2.111

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

c. The test statistic is

t= (x1`-x2`) /√s1²/n1 + s2²/n2

t= (21-27) /√5.6²/10+ 4.3²/14

t= -6/2.111

t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

5 0

3 years ago