Need help It is working with functions and I need assistance

Mathematics

1 answer:

Ainat [17]3 years ago

5 0

Answer: its 4. Look at the graph, x is horizontal and y is vertical (first go sideways then go up.) if coordinates are x,y then first go sideways when looking for x. they give you x. 7. then we go up. the graph intersects 7 at 4, therefore the coords are 7, 4 and your y coordinate is 4.

Step-by-step explanation:

You might be interested in

I need help on this how do you find the third measure

Goryan [66]

The sum of the three angles of a triangle is always 180°. Tha is
(first measure) + (second measure) + (third measure) = 180°

Subtract the first two terms of this equation and you have
(third measure) = 180° - (first measure) - (second measure)

a. third measure = 180° - 45° - 55° = 80°

b. third measure = 180° - 50° - 40° = 90°

c. third measure = 180° - 110° - 35° = 35°

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:

$X \sim N(13.35,0.12)$