Write a quadratic function that reflects the x axis, stretches by a factor of 5, and shifts left 6 and up 3

A 2-column table with 5 rows. Column 1 is labeled Number of chores, x with entries 2, 4, 6, 8, 10. Column 2 is labeled Days, y w

Alexandra [31]

Answer:

C: 1/2

Step-by-step explanation:

6 0

2 years ago

Ví dụ 1.5. Một người đi mua hàng 3 lần. Xác suất lần đầu mua được hàng tốt là 0,7.

vodka [1.7K]

Answer:

a) 0.50575,

b) 0.042

Step-by-step explanation:

Example 1.5. A person goes shopping 3 times. The probability of buying a good product for the first time is 0.7.

If the first time you can buy good products, the next time you can buy good products is 0.85; (I interpret this as, if you buy a good product, then the next time you buy a good product is 0.85).

And if the last time I bought a bad product, the next time I bought a good one is 0.6. Calculate the probability that:

a) All three times the person bought good goods.

P(Good on 1st shopping event AND Good on 2nd shopping event AND Good on 3rd shopping event) =

P(Good on 1st shopping event) *P(Good on 2nd shopping event | Good on 1st shopping event) * P(Good on 3rd shopping event | 1st and 2nd shopping events yield Good) =

(0.7)(0.85)(0.85) =

0.50575

b) Only the second time that person buys a bad product.

P(Good on 1st shopping event AND Bad on 2nd shopping event AND Good on 3rd shopping event) =

P(Good on 1st shopping event) *P(Bad on 2nd shopping event | Good on 1st shopping event) * P(Good on 3rd shopping event | 1st is Good and 2nd is Bad shopping events) =

(0.7)(1-0.85)(1-0.6) =

(0.7)(0.15)(0.4) =

0.042

5 0

3 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

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".

Part a

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

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

5 0

3 years ago

WILL GIVE BRAINLIEST

Olegator [25]

The answer is A.
We can first eliminate D since it uses these (<, >) signs and the lines are shaded, indicating the points on those lines are solutions.
We can also eliminate C because the y intercept in C’s lines is 2, while in the graph, they are both 3.
Finally, we can look at both inequalities on the graph and see that the shaded areas are both underneath the line. This means that y is less than the equation for the line, eliminating B
So, the answer is A
Hope this made sense!!

7 0

3 years ago

Alexa and Brent are each working on a 30-question assignment. Alexa completed 20% of the questions. Brent completed 30% of the q

Mandarinka [93]

So 20% of 30 is 6
30% * 30 is 9
Alexa completed 6 of 30 questions
Brent completed 9 of 30 questions

Brent completed 3 more questions than Alexa

Hope this helps :)

6 0

3 years ago