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Sedbober [7]
2 years ago
11

Which of these is a correct expansion of (4x - 2)(2x2 + 3)?

Mathematics
1 answer:
Neporo4naja [7]2 years ago
5 0

Answer:

a correct expansion is B

You might be interested in
What is the value of x when -25(x - 4) = -55(x - 10) ?
Free_Kalibri [48]

Simplify both sides of the equation

-25(x-4)=-55(x-10)

(-25)(x) + (-25)(-4)=(-55)(x) + (-55) (-10)

Distribute

-25x+100=-55x+550

Add 55x  to both sides

-25x+100+55x=-55x+550+55x

30x+100-100=550-100

subtract 100 from both sides

30x+100-100=550-100

30x=450

divide both sides by 30

30x/30=450/30

x= 15

The value of x is 15


I hope that's help you my friend !

5 0
3 years ago
Read 2 more answers
Find the missing Side of the triangle
sammy [17]

Answer:

2√15

Step-by-step explanation:

Use the Pythagorean theorem.

2² + x² = 8²

x² + 4 = 64

x² = 60

x² = 4 * 15

x = 2√15

3 0
2 years ago
Math math math math math <br> i need help<br> tysm :) &lt;3
stiv31 [10]

Answer:

C. x^14

Step-by-step explanation:

(x^9)^0 is 1

(x^7)^2 is x^14

x^14*1 is x^14

The Answer is C. x^14

Hope this helps!

5 0
2 years ago
Ninety-one percent of products come off the line within product specifications. Your quality control department selects 15 produ
Allisa [31]

Answer:

Probability of stopping the machine when X < 9 is 0.0002

Probability of stopping the machine when X < 10 is 0.0013

Probability of stopping the machine when X < 11 is 0.0082

Probability of stopping the machine when X < 12 is 0.0399

Step-by-step explanation:

There is a random binomial variable X that represents the number of units come off the line within product specifications in a review of n Bernoulli-type trials with probability of success 0.91. Therefore, the model is {15 \choose x} (0.91) ^ {x} (0.09) ^ {(15-x)}. So:

P (X < 9) = 1 - P (X \geq 9) = 1 - [{15 \choose 9} (0.91)^{9}(0.09)^{6}+...+{ 15 \choose 15}(0.91)^{15}(0.09)^{0}] = 0.0002

P (X < 10) = 1 - P (X \geq 10) = 1 - [{15 \choose 10}(0.91)^{10}(0.09)^{5}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0013

P (X < 11) = 1 - P (X \geq 11) = 1 - [{15 \choose 11}(0.91)^{11}(0.09)^{4}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0082

P (X < 12) = 1- P (X \geq 12) = 1 - [{15 \choose 12}(0.91)^{12}(0.09)^{3}+...+{15 \choose 15} (0.91)^{15}(0.09)^{0}] = 0.0399

Probability of stopping the machine when X < 9 is 0.0002

Probability of stopping the machine when X < 10 is 0.0013

Probability of stopping the machine when X < 11 is 0.0082

Probability of stopping the machine when X < 12 is 0.0399

8 0
2 years ago
You invested $19,000 in two accounts paying 4% and 8% annual interest, respectively.
saveliy_v [14]

Answer:

$9000 at 4$

and

$10000 at 8%

Step-by-step explanation:

Let's assume that "x" is the amount deposited in the 4% account and "y" is the amount deposited in the 8% account.

Recall the formula for interest as : I= P * R *t

where I is the interest, R is the annual rate of interest and t is the number of years.

Since there are two investments, we need to add both interests at the end of the one year: I1 = x (0.04) (1) = 0.04 x   and   I2 = y (0.08) (1) = 0.08 y

Total Interest = Interest (from the 4% account) + Interest (from the 8% account)

Total Interest = $1160 = 0.04 x +  0.08 y

we also know that the total invested (x + y) adds to $19,000, that is:

$19,000 = x + y

Then we can solve these system of two equations by substitution, for example solving for y in the second equation and using the y substitution in the first equation;

y = 19000 - x

1160 = 0.04 x + 0.08 (19000 - x)

1160 = 0.04 x + 1520 - 0.08 x

0.08 x - 0.04 x = 1520 - 1160

0.04 x = 360

x = 360/0.04 = $9000

Then the other investment was : y = $19000 - $9000 = $10000

6 0
2 years ago
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