Mr. Ruiz drove 205 miles in 5 hours on Saturday and 180 miles in 4 hours on Sunday. What was his average speed, in miles per hour, for the two days

Answers

Answer 1
Answer:

Answer:

In miles per hour was 34 mph

Step-by-step explanation:

205/5 = 41

180/4 = 27

to find the average you add both speeds then divide by how many there are.

41 + 27 = 68

68/2 = 34


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Solve for x. x+27=−12

Answers

x + 27 = -12
subtract 27 from both sides
x = -12-27
x = -39
-39 is the correct answer

each day a bobcat drinks about 3 times as much water as a canada goose drinks. how much water can a bobcat drink in ① day?

Answers

56 gallons
I think..
Hope this will be helpful for you
the formula for this is b= bobcat and g=goose so if the bobcat is three times as much as the goose then b=3g now if you know b or g you can fill it in to find the other.

Nancy hears a report that the current national average price of gasoline is $3.14. She averages the prices of stations near her home.she finds the average price of gas to be $2.95. Why are the averages different?

Answers

Answer:

Nancy's ambient locality might be having more suppliers, higher supply of Gasoline

Step-by-step explanation:

National average price of gasoline = $3.14

Stations near Nancy's home have an average price = $2.95

Gasoline price near Nancy's house is lesser than national average price level. Number of gasoline stations near her house might be more. It implies increase in its supply. This leads to competition among sellers & reduces the price level.

An ordinary regression model that treats the response Y as normally distributed is a special case of a GLM, with normal random component and identity link function. 1. With a GLM. Y does not need to have a normal distribution and one can model a function of the mean of Y instead of just the mean itself, but In order to get the maximum likelihood estimates the variance of Y must be constant at all values of predictors. ii. The Pearson residual e_inty_i-muhatiysqrt(muhat_) for a GM has a large-sample standard normal distribution (a)) True False, (w) True; (b)) True 00 True True (co False 0 false, Oll) False; IdFalse 00 True (1) False;

Answers

An ordinary Regression model that treats the response Y is (a) True False, (w) True

What is Regression?

A statistical method called regression links a dependent variable to one or more independent (explanatory) variables.

A regression model can demonstrate whether changes in one or more of the explanatory variables are related to changes in the dependent variable.

A) Models for numerical response variable, like ANOVA and linear regression are special cases of GLMs

for these model the following holds

1. Random component has a normal distribution

2. Systematic component α+β₁x₁+β₂x₂+...........βₓxₓ

3. link function = identity (g(µ)=µ)

GLMs can generalise these models with response Y as normally distributed, hence the statement is True

B) With a GLM. Y does not need to have a normal distribution and one can model a function of the mean of Y instead of just the mean itself. but in order to get ML estimates the variance of Y must be small. This small variance of Y is the reason for ML estimator to be the best one. hence the statement is false.

An ordinary Regression model that treats the response Y is (a) True False, (w) True

To learn more about Regression Visit:

brainly.com/question/14313391

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Stuck on letter b, please help

Answers

Answer:  f(3)=10

Step-by-step explanation:

In order to calculate f(3) you need to substitute 3 for x in function

f(x)=3x +1 :

f(3)=( 3x3) + 1

f(3)= 9+1

f(3)=10

What sample size, including the 20 observations in the initial study, would be necessary to have a confidence of 95.44 percent that the observed time was within 4 percent of the true value?

Answers

Completed question:

An initial time study resulted in an average observed time of 2.2 minutes per cycle, and a standard deviation of .3 minutes per cycle. The performance rating was 1.20. What sample size, including the 20 observations in the initial study, would be necessary to have a confidence of 95.44 percent that the observed time was within 4 percent of the true value?

Answer:

47

Step-by-step explanation:

When doing a statistic study, a sample of the total amount must be taken. This sample must be done randomly, and, to be successful, the sample size (n) must be determined, by:

n = ((Z_(\alpha/2)*S )/(E))^2

Where Z(α/2) is the value of the standard normal variable associated with the confidence, S is the standard deviation, and E is the precision. The confidence indicates if the study would have the same result if it would be done several times. For a confidence of 95.44, Z(α/2) = 2.

The standard deviation indicates how much of the products deviate from the ideal value, and the precision indicates how much the result can deviate from the ideal. So, if it may vary 4% of the true value (2.2), thus E = 0.04*2.2 = 0.088.

n = [(2*0.3)/0.088]²

n = 46.48

n = 47 observations.