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Nat2105 [25]
3 years ago
12

Solve the following equation: 3x2 = -12 a. I = 2 b. r=-2 c. No solution d. I= +2

Mathematics
2 answers:
ella [17]3 years ago
4 0
Your is is no solution
Delicious77 [7]3 years ago
3 0

Answer:

c

Step-by-step explanation:

Given

3x² = - 12 ( divide both sides by 3 )

x² = - 4 ( take the square root of both sides )

x = ± \sqrt{-4} ← has no real solutions → c

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Carol is cross country skiing the table shows the distance traveled after various numbers of minutes what is the rate of change
MatroZZZ [7]

Answer:

Option D 3/16

Step-by-step explanation:

Carol is cross-country skiing.

With the help of the given table we have to calculate the rate of change.

If we draw a graph for distance traveled on y-axis and tins at x-axis we find two points, ( 2, 1/6) and (3, 17/48).

Then slope of the line connecting these points will be the rate of change.

8 0
3 years ago
Factor completely, then place the factors in the proper location on the grid. 25a2 +9b2 + 30ab
sergij07 [2.7K]

Answer:

(5a+3b)(5a+3b) or SQ(5a+3b)

Step-by-step explanation:

Now you can plot by referring to the above factors

8 0
3 years ago
Ms. Edwards is redecorating her office. She has a choice of 6 colors of paint, 3 kinds of curtains, and 4 colors of carpet. How
MariettaO [177]
6 * 3 * 4 = 72 ways :)
7 0
4 years ago
Hi whats <br> u = x - k solve for x
kykrilka [37]

Answer:

x=u+k

Step-by-step explanation:

4 0
2 years ago
Read 2 more answers
Assume that foot lengths of women are normally distributed with a mean of 9.6 in and a standard deviation of 0.5 in.a. Find the
Makovka662 [10]

Answer:

a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

\mu = 9.6, \sigma = 0.5.

a. Find the probability that a randomly selected woman has a foot length less than 10.0 in

This probability is the pvalue of Z when X = 10.

Z = \frac{X - \mu}{\sigma}

Z = \frac{10 - 9.6}{0.5}

Z = 0.8

Z = 0.8 has a pvalue of 0.7881.

So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.

b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.

When X = 10, Z has a pvalue of 0.7881.

For X = 8:

Z = \frac{X - \mu}{\sigma}

Z = \frac{8 - 9.6}{0.5}

Z = -3.2

Z = -3.2 has a pvalue of 0.0007.

So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.

c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.

Now we have n = 25, s = \frac{0.5}{\sqrt{25}} = 0.1.

This probability is 1 subtracted by the pvalue of Z when X = 9.8. So:

Z = \frac{X - \mu}{s}

Z = \frac{9.8 - 9.6}{0.1}

Z = 2

Z = 2 has a pvalue of 0.9772.

There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.

5 0
3 years ago
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