A map has a scale of 6 in : 26 mi. If Clayton and Clinton are 52 mi apart, then they are how far apart on the map?

Two kinds of tickets to an outdoor concert were sold: lawn tickets and seat tickets. Less than 400 tickets in total were sold. L

miskamm [114]

Answer:

Tickets to sit on the bench (b) = 150

Tickets to sit on the lawn (l) = 200

Step-by-step explanation:

Tickets to sit on a bench (b)cost $75 each.

Tickets to sit on the lawn (l) cost $40 each

350 tickets had been sold

$19,250 had been raised through tickets sales

This forms a simultaneous equation:

b + l = 350 ... (i)

75b + 40l = 19,250 ... (ii)

Multiplying (i) by 40 and (ii) by 1 we get;

40b + 40l = 14,000 ... (i)

75b + 40l = 19,250 ... (ii)

Subtracting (ii) - (i) we get;

35b = 5250

b = 5250 ÷ 35 = 150

7 0

3 years ago

What is the inverse of this function? f(x)=x−7

viva [34]

Answer:

Step-by-step explanation:

Replace f(x) with y. y = x − 7

Interchange the variables. x = y − 7

Solve for y. y = x + 7

Solve for y and replace with f^−1 (x). f^−1(x) = x + 7

I hope this helps have a great day :)

4 0

2 years ago

Worth 2 marks and I need it urgently pls

melisa1 [442]

4c6d3 x 3c4d

4 x 3=12
12c
you add the powers so 6+4=10
12c10
d3 x d=d3
12c10d3

that is the answer i think

4 0

2 years ago

Can someone answer this math problem!

mr_godi [17]

Answer:

b i think i am wrong idk

Step-by-step explanation:

7 0

2 years ago

SAT scores are normed so that, in any year, the mean of the verbal or math test should be 500 and the standard deviation 100. as

vovangra [49]

Answer:

a) $P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

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

b) $P(400$

$P(-1$

c) $z=-0.842$

And if we solve for a we got

$a=500 -0.842*100=415.8$

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.

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

$X \sim N(500,100)$

Where $\mu=500$ and $\sigma=100$

We are interested on this probability

$P(X>625)$

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:

$P(X>625)=P(\frac{X-\mu}{\sigma}>\frac{625-\mu}{\sigma})=P(Z>\frac{625-500}{100})=P(Z>1.25)$

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

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

Part b

We are interested on this probability

$P(400$

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:

$P(400$

And we can find this probability with this difference:

$P(-1$

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

$P(-1$

Part c

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

$P(X>a)=0.8$ (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 0.2 of the area on the left and 0.8 of the area on the right it's z=-0.842. On this case P(Z<-0.842)=0.2 and P(Z>-0.842)=0.8

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:

$z=-0.842$

And if we solve for a we got

$a=500 -0.842*100=415.8$

So the value of height that separates the bottom 20% of data from the top 80% is 415.8.

8 0

2 years ago