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2 years ago

Please help me out!!!!!!!!

Mathematics

1 answer:

Aloiza [94]2 years ago

6 0

I believe that the answer is 70. I hope I helped! The reason I say this is because BD is half of AE and 35+35=70 :)

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PLZ HELP ME WITH THIS!! 5/2 c = 8 1/3 *

Ivahew [28]

Answer:

3 1/3

Step-by-step explanation:

x=10/3

7 0

3 years ago

Solve x²+5x+6=0 by completing the square

Brums [2.3K]

Answer:

x = -2

x= -3

Step-by-step explanation:

x 2+5x+6=0

To solve the equation, factor x^2+5x+6 using formula x^2+(a+b)x+ab=(x+a)(x+b). To find a and b, set up a system to be solved.

a+b=5

ab=6

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

1,6

2,3

Calculate the sum for each pair.

1+6=7

2+3=5

The solution is the pair that gives sum 5.

a=2

b=3

Rewrite factored expression (x+a)(x+b) using the obtained values.

(x+2)(x+3)

To find equation solutions, solve x+2=0 and x+3=0.

x=−2

x=−3

4 0

3 years ago

The distribution of weights of potato chip bags filled off a production line is unknown. However, the mean is m=13.35 OZs and th

Nikitich [7]

Answer:

a) $\mu_{\bar X} =13.35$

e.none of the above

b) $\sigma_{\bar X}= \frac{0.12}{\sqrt{36}}= 0.02$

a.0.0200

c) $z = \frac{13.32-13.35}{\frac{0.12}{\sqrt{36}}}= -1.5$

$P(\bar X< 13.32)= P(z$

a.0.0668

d) $z = \frac{13.30-13.35}{\frac{0.12}{\sqrt{36}}}= -2.5$

$z = \frac{13.36-13.35}{\frac{0.12}{\sqrt{36}}}= 0.5$

$P(13.30$

c.0.6853

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(13.35,0.12)$