Kamil has 72 baseball cards. He wants to put the cards

A consensus forecast is the average of a large number of individual analysts' forecasts. Suppose the individual forecasts for a

zaharov [31]

Answer:

a) $P(X>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)$

And we can find this probability using the complement rule:

$P(z>-1.25)=1-P(z$

b) $P(X$

And we can find this probability uing the normal standard table:

$P(z$

c) $P(3.5$

And we can find this probability with this difference:

$P(-1.25$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-1.25$

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 variable of interest of a population, and for this case we know the distribution for X is given by:

$X \sim N(5,1.2)$

Where $\mu=5$ and $\sigma=1.2$

We are interested on this probability

$P(X>3.5)$

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:

$P(X>3.5)=P(\frac{X-\mu}{\sigma}>\frac{3.5-\mu}{\sigma})=P(Z>\frac{3.5-5}{1.2})=P(z>-1.25)$

And we can find this probability using the complement rule:

$P(z>-1.25)=1-P(z$

Part b

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 uing the normal standard table:

$P(z$

Part c

$P(3.5$

And we can find this probability with this difference:

$P(-1.25$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-1.25$

3 0

3 years ago

What is the answer to this equation -14=d/15

expeople1 [14]

Answer:

d=-210

Step-by-step explanation:

multiply 15 to both sides

-14*15 =-210

the 15 on the side with d cancels

Now you have d=-210

Hope this helps!

3 0

3 years ago

What is 1 and 1/4% of 200?

arsen [322]

5/4 • 200/1, 1000/4
which is 250

250 is 1 1/4 of 200

4 0

3 years ago

Simplify the expression

Molodets [167]

Answer:

8x^8/3 y^4 - The first option

Step-by-step explanation:

The first thing we need to do is the exponent outside the bracket and leave the 8 till last, because exponents always come before multiplying coefficients. When an term with an exponent is multiplied by another exponent outside a bracket, the exponents of both terms are multiplied by the exponent outside the bracket.

This means that the expression is now:

8(x^2*4/3 y^3*4/3)

First we can so the x term. The x term already has an exponent of 2, so the 2 is multiplied by the 4/3 exponent outside the bracket. 2*4/3 = 8/3, so the x term is now: x^8/3

The same happens to the y term: 3*4/3 simplifies to 4, so the y term is now y^4.

So now our expression is:

8(x^8/3 y^4)

Now the 8 outside the bracket simply multiplies on to the whole term so we finish with:

8x^8/3 y^4 - The first option.

Hope this helped!

7 0

3 years ago

Really need an answer to this please help.

Alex

Answer:

a

Step-by-step explanation:

$\sqrt[4]{144a^{12}b^{3}} = \sqrt[4]{4^{2}*3^{2}a^{12}b^{3}}=\\= \sqrt[4]{2^{4}*3^{2}a^{12}b^{3}}=2a^{3}\sqrt[4]{3^{2}b^{3}} =\\}=2a^{3}\sqrt[4]{9b^{3}}$

7 0

3 years ago