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3 years ago

Find the solutions to the equation below check all that apply

Mathematics

2 answers:

jek_recluse [69]3 years ago

7 0

Answer:

C. x=-1/2 and F. x=-3

Step-by-step explanation:

We can solve this by factoring. We will look for factors of a*c, 2*3, that sum to b, 7. Factors of 6 are:

1 and 6; these sum to 7.

2 and 3; these do not sum to 7.

We will use 1 and 6. We will split bx, 7x, into 1x and 6x:

2x²+1x+6x+3 = 0

We will group the first two and the last two terms together:

(2x²+1x)+(6x+3) = 0

We will factor out the GCF of each group. The GCF of the first group is 1x:

1x(2x+1)+(6x+3) = 0

The GCF of the second group is 3:

1x(2x+1)+3(2x+1) = 0

The GCF of the two terms remaining is (2x+1); factoring this out,

(2x+1)(1x+3) = 0

Using the zero product property,

2x+1 = 0 or 1x+3 = 0

To solve the first equation, subtract 1 from each side:

2x+1-1 = 0-1

2x = -1

Divide both sides by 2:

2x/2 = -1/2

x = -1/2

To solve the second equation, subtract 3 from each side:

1x+3-3 = 0-3

x = -3

Harman [31]3 years ago

4 0

2x^2+7x+3=0: x=-1/2 and x=-3 c and F

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a teacher has given a test to a group of students. the set of scores is: 78, 79, 79, 81, and 83. find the variance for this samp

Delvig [45]

Based on the data set, the variance for this sample of tests is 4

<h3>How to find the variance for this sample of tests?</h3>

The sample of test is given as:

78, 79, 79, 81, and 83.

Calculate the mean, using:

Mean = Sum/Count

So, we have

Mean = (78 + 79 + 79 + 81 + 83)/5

Evaluate

Mean = 80

The variance is calculated as

$\sigma^2 = \frac{\sum(x - \bar x)^2}{n-1}$

This gives

$\sigma^2$ = [(78 - 80)^2 + (79 - 80)^2 + (79- 80)^2 + (81- 80)^2 + (83- 80)^2)]/[5 - 1]

Evaluate the numerator and the denominator.

So, we have:

$\sigma^2$ =16/4

Evaluate the quotient

$\sigma^2$ = 4

Hence, the variance for this sample of tests is 4

Read more about variance at:

brainly.com/question/15858152

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7 0

2 years ago

The probability that the noise level of a wide-band amplifier will exceed 2 dB is 0.05 independently of every other amplifier. C

snow_lady [41]

Answer:

0.980

Step-by-step explanation:

The probability that the noise level of a wide-band amplifier will exceed 2 dB is 0.05

So, probability of success = 0.05

Probability of failure = 1-0.05=0.95

There are 12 amplifiers

We are supposed to find the probability that at most two will exceed 2dB.

We will use binomial distribution

Formula : $P(X=r)=^nC_r p^r q ^{n-r}$

p = 0.05

q = 0.95

n = 12

We are supposed to find the probability that at most two will exceed 2dB.

So, $P(X\leq 2)=P(X=0)+P(X=1)+P(X=2)$

$P(X\leq 2)=^{12}C_0 P(0.05)^0 (0.95)^{12-0}+^{12}C_1 P(0.05)^1(0.95)^{12-1}+^{12}C_2 P(0.05)^2 (0.95)^{12-2}$

$P(X\leq 2)=\frac{12!}{0!(12-0)!} (0.05)^0 (0.95)^{12-0}+\frac{12!}{1!(12-1)!}(0.05)^1(0.95)^{12-1}+\frac{12!}{2!(12-2)!} (0.05)^2 (0.95)^{12-2}$

$P(X\leq 2)=0.980$

Hence the probability that at most two will exceed 2dB is 0.980

7 0

3 years ago

Round to the nearest hundredth 42.052

konstantin123 [22]

Answer:

To round 42.052 to the nearest hundredth consider the thousandths' value of 42.052, which is 2 and less than 5. Therefore, the hundredths' value of 42.052 remains 5.

5 0

3 years ago

Read 2 more answers

Can someone help me with this

rusak2 [61]

Hello There!

First let me say this. I did not like how this question was worded I am a tutor in math and I had to read this problem over and over 5 times. It was confusing but I managed to figure it out. The answer is 61%

My work is shown on paper

4 0

3 years ago

Can somebody please help me?

9966 [12]

28+28+28= 84

28 x 28 x 28 = 21952

hope it helps!:)

4 0

3 years ago