A total of 634 tickets were sold for the school play. They were either adult tickets or student tickets. There were 66 fewer students tickets sold than adult tickets. Him many adult tickets were sold?

Answers

Answer 1
Answer:  A total of 634 tickets were sold for the school play. They were either adult tickets or student tickets. There were 66 fewer student tickets sold than adult tickets. How many adult tickets were sold?
So say adult sold would be x, then student sold would be x-66
add them

x+x-66=634 solve for x
2x=700 divide each side by 2
x=350
so 350 adult and 350-66=284 student
Answer 2
Answer: x= student tickets
y= adult tickets 

x+y=634
y=x-66

x+y=634 x-y=66   2x=700 x=350 

350+y=634  y=284

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If the level of significance of a hypothesis test is raised from .01 to .05, the probability of a Type II error a. will not change. b. will increase. c. will also increase from .01 to .05. d. will decrease.

Answers

Answer:

d. Decrease

Step-by-step explanation:

A Type II error is when we fail to reject a false null hypothesis. Higher values of α make it easier to reject the null hypothesis, so choosing higher values for α can reduce the probability of a Type II error.

The consequence here is that if the null hypothesis is true, increasing α makes it more likely that we commit a Type I error (rejecting a true null hypothesis).

So using lower values of α can increase the probability of a Type II error.

Final answer:

Raising the level of significance in a hypothesis test from .01 to .05 would decrease the probability of making a Type II error. This is because as we become more accepting of risk in making a Type I error, we simultaneously reduce the risk of making a Type II error.

Explanation:

The level of significance in a hypothesis test is the probability that we are willing to accept for incorrectly rejecting the null hypothesis or making a Type I error. If the level of significance is raised, there is a higher chance we incorrectly reject the null hypothesis, increasing the chances of a Type I error. However, this also has an effect on the probability of committing a Type II error, which is to incorrectly accept the null hypothesis.

Specifically, when the level of significance of a hypothesis test is raised from .01 to .05, the probability of a Type II error (option b) will decrease. The reason for this is that increasing the level of significance or alpha means we are more likely to reject the null hypothesis. As we are more accepting of risk in terms of making a Type I error, we are less likely to make a Type II error, as the two error types often move in opposite directions. Thus, the answer to your question is d. The probability of a Type II error will decrease if the significance level is raised from .01 to .05.

Learn more about Hypothesis Testing here:

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If x - 10 = -15, then find the value of 5x - 30 *

Answers

Answer:

The value is -55

In the equation x-10=-15, x is equal to -5.

Therefore, the value of x is plugged into the second equation and the value found is -55.

Answer:

x = -55

Step-by-step explanation:

We know that x - 10 = -15 and we want to find the value of 5x - 30 but our first step would be to find the value x so then we can can substitute that back into the expression so

x - 10 = -15

⇔ Add -10 to both sides to isolate x

x = -5

So now we know that value of 'x' is -5 we can substitute it back into the expression 5x - 30 so

5x - 30

→ Substitute x = -5 back into it

5 × -5 - 30

→ Simplify

-25 - 30 = -55

If x - 10 = -15, then find the value of 5x - 30 the value of x is -55

In determining automobile-mileage ratings, it was found that the mpg (X) for a certain model is normally distributed, with a mean of 33 mpg and a standard deviation of 1.7 mpg. Find the following:__________. a. P(X<30)
b. P(28c. P(X>35)
d. P(X>31)
e. the mileage rating that the upper 5% of cars achieve.

Answers

The upper 5% of cars have a mileage rating of 35.805 mpg

What is z score?

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

Given;  mean of 33 mpg and a standard deviation of 1.7

a) For < 30:

z = (30 - 33)/1.7 = -1.76

P(x < 30) = P(z < -1.76) = 1 - 0.8413 = 0.0392

b) For < 28:

z = (28 - 33)/1.7 = -2.94

P(x < 28) = P(z < -2.94) = 0.0016

c) For > 35:

z = (35 - 33)/1.7 = 1.18

P(x > 35) = P(z > 1.18) = 1 - P(z < 1.18) = 1 - 0.8810 = 0.119

d) For > 31:

z = (31 - 33)/1.7 = -1.18

P(x > 31) = P(z > -1.18) = 1 - P(z < -1.18) = 0.8810

e) The  upper 5% of cars achieve have a z score of 1.65, hence:

1.65 = (x - 33)/1.7

x = 35.805 mpg

The upper 5% of cars have a mileage rating of 35.805 mpg

Find out more on z score at: brainly.com/question/25638875

Answer:

a) P(X < 30) = 0.0392.

b) P(28 < X < 32) = 0.2760

c) P(X > 35) = 0.1190

d) P(X > 31) = 0.8810

e) At least 35.7965 mpg

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question, we have that:

\mu = 33, \sigma = 1.7

a. P(X<30)

This is the pvalue of Z when X = 30. So

Z = (X - \mu)/(\sigma)

Z = (30 - 33)/(1.7)

Z = -1.76

Z = -1.76 has a pvalue of 0.0392.

Then

P(X < 30) = 0.0392.

b) P(28 < X < 32)

This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 28. So

X = 32

Z = (X - \mu)/(\sigma)

Z = (32 - 33)/(1.7)

Z = -0.59

Z = -0.59 has a pvalue of 0.2776.

X = 28

Z = (X - \mu)/(\sigma)

Z = (28 - 33)/(1.7)

Z = -2.94

Z = -2.94 has a pvalue of 0.0016.

0.2776 - 0.0016 = 0.2760.

So

P(28 < X < 32) = 0.2760

c) P(X>35)

This is 1 subtracted by the pvalue of Z when X = 35. So

Z = (X - \mu)/(\sigma)

Z = (35 - 33)/(1.7)

Z = 1.18

Z = 1.18 has a pvalue of 0.8810.

1 - 0.8810 = 0.1190

So

P(X > 35) = 0.1190

d. P(X>31)

This is 1 subtracted by the pvalue of Z when X = 31. So

Z = (X - \mu)/(\sigma)

Z = (31 - 33)/(1.7)

Z = -1.18

Z = -1.18 has a pvalue of 0.1190.

1 - 0.1190 = 0.8810

So

P(X > 31) = 0.8810

e. the mileage rating that the upper 5% of cars achieve.

At least the 95th percentile.

The 95th percentile is X when Z has a pvalue of 0.95. So it is X when Z = 1.645. Then

Z = (X - \mu)/(\sigma)

1.645 = (X - 33)/(1.7)

X - 33 = 1.645*1.7

X = 35.7965

At least 35.7965 mpg

Kayla had 40 minutes left beforebedtime. Then she worked on a
painting for 20 minutes, took a shower
for 8 minutes, and read a book for 10
minutes. How many more minutes are
left before Kayla's bedtime?

Answers

Answer:

2 minutes are left

Step-by-step explanation:

1. Start with 40

2. Subtract 20 from 40 (40-20) to get 20 minutes.

3. Then take 8 minutes from 20 (20-8) to get 12 minutes

4. Lastly, take 10 minutes away from 12 (12-10) to get 2 minutes left.

5. This leaves Kayla with 2 minutes before bedtime.

Help to solve   y'''-y''-4y'+4y=5-e^x+e^2x

Answers

By solving the equation "y'''-y''-4y'+4y=5-e^x+e^2x" we get "y = Ae^x+Be^(2e)+Ce^(-2x)+(5)/(4)+(xe^x)/(3) +(xe^(2x))/(4)".

As we know the auxiliary equations,

  • m^3-m^2-4m+4=0
  • (m-1)(m-2)(m+2) =0

here,

m = 1, 2, -2

Thus,

The general equation will be:

y = Ae^x+Be^(2x)+Ce^(-2x)

Particularly,

  • 5 = (5)/(4)
  • e^x= x((1)/(1-4)e^x ) = -(xe^x)/(3)
  • e^(2x) = x((e^(2x))/((2+2)(2-1)) )= (xe^(2x))/(4)

hence,

The complete equation will be:

y = Ae^x+Be^(2e)+Ce^(-2x)+(5)/(4)+(xe^x)/(3) +(xe^(2x))/(4)

Thus the above answer is right.

Learn more:

brainly.com/question/22449567

Answer:

Step-by-step explanation:

First of all write the auxialary equation as

m^3-m^2-4m+4 =0\n(m-1)(m-2)(m+2)=0

m=1,2,-2

Hence general solution is

y=Ae^x+Be^(2x) +Ce^(-2x)

Particular solution of 5 is

(5)/(4)

Particular solution of e^x is

x((1)/(1-4) e^x =(-xe^x)/(3)

Particular solution of e^2x is

x(e^(2x) )/((2+2)(2-1)) =(xe^(2x) )/(4)

Together full solution is

y=Ae^x+Be^(2x) +Ce^(-2x)+(5)/(4) +(xe^x)/(3) +(xe^(2x) )/(4)

What is the lateral surface area of this cylinder? r = 7ft h = 16ft Answers:

702.96 sq ft
703.36 sq ft
703.96 sq ft
704.36 sq ft​

Answers

Answer:

703.36 sq ft

Step-by-step explanation:

The lateral area is the product of the circumference and the height.

LA = 2π·r·h = 2·3.14·(7 ft)(16 ft) = 703.36 ft²

_____

Comment on the answer choices

Please note this is not the actual area, which rounds to 703.72 ft². The above value is obtained only by using an inappropriate value for π. The 5-significant-digit answer is not supported by a 3-significant-digit value for π. More appropriate would be π≈3.1416.