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3 years ago

A blueprint for a house has a scale factor = 10. A wall in the blueprint is 8 in. What is the length of the actual wall?

Mathematics

1 answer:

oksian1 [2.3K]3 years ago

7 0

Answer: 6.67 feet (check to see if you made a typo)

Step-by-step explanation:

if the scale factor is ten than we multiply 8 by 10

8 * 10 = 80

important to remember this is 80 INCHES not 80 FEET

to convert 80 inches into feet, we divide by 12 because there are 12 inches in a foot

80/12 = 6.67 feet

check to make sure that the first option is actually 6.67 inches because it would make more sense for it to be in feet

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ASAP PLEASE (this is like five questions in one) GIVING 15 POINTS!!!

valentinak56 [21]

Answer:

21√2/2

Step-by-step explanation:

For all problems, the ratios of the lengths of the sides are:

30-60-90 triangle 1 : √3 : 2

45-45-90 triangle 1 : √2 : √2

Problem 3:

30-60-90 triangle at left

Hypotenuse = 11

Short leg = 11/2

Long leg = 11/2 × √3

45-45-90 triangle at right

x = 11/2 × √3 × √2 = 11√6/2

Answer: D

Problem 4:

30-60-90 triangle at top

Hypotenuse = 7√3

Short leg = 7√3/2

Long leg = 7√3/2 × √3 = 7 × 3/2 = 21/2

45-45-90 triangle at right

x = 21/2 × √2 = 21√2/2

Answer: 21√2/2

Problem 7:

30-60-90 triangle

Long leg = 4√3

Short leg = y = 4√2/2 = 2√2

Answer: C.

Problem 6:

30-60-90 triangle at left

Hypotenuse = 12√6

Short leg = 12√6/2 = 6√6

Long leg = 6√6 × √3 = 6√2√3√3 = 18√2

45-45-90 triangle at right

x = 18√2/√2 = 18

Answer: C.

Problem 8:

45-45-90 triangle

Hypotenuse = 2

Leg = y = 2/√2 = 2√2/(√2√2) = √2

Answer: C.

5 0

2 years ago

What the decimal of 7/9

Mashutka [201]

Divide 7 and 9

7/9 = 0.77 (repeating)

0.77 (repeating) rounded would be 0.78

0.78 is your answer

hope this helps

5 0

4 years ago

The data below represent the number of days absent, x, and the final grade, y, for a sample of college students at a large uni

Zina [86]

Answer:

The equation of the regression line is: $y~=~88.518 ~-~ 3.068 \cdot x$

Step-by-step explanation:

Linear regression is a linear approach to modelling the relationship between a dependent variable and one or more independent variables.

Let X be the independent variable and Y be the dependent variable. We will define a linear relationship between these two variables as follows:

$Y=bX+a$

We have the the following data:

$\begin{array}{c|cccccccccc}No. \:of \:absences,\:x&0&1&2&3&4&5&6&7&8&9\\Final \:grade, y&88.5&85.7&82.8&80.3&77.4&73.1&63.6&68.1&65.2&62.4\end{array}$

To find the line of best fit for the points, follow these steps:

Step 1: Find $X\cdot Y$ and $X\cdot X$ as it was done in the table.

Step 2: Find the sum of every column:

$\sum{X} = 45 ~,~ \sum{Y} = 747.1 ~,~ \sum{X \cdot Y} = 3108.8 ~,~ \sum{X^2} = 285$

Step 3: Use the following equations to find intercept a and slope b:

$\begin{aligned} a &= \frac{\sum{Y} \cdot \sum{X^2} - \sum{X} \cdot \sum{XY} }{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 747.1 \cdot 285 - 45 \cdot 3108.8}{ 10 \cdot 285 - 45^2} \approx 88.518 \\ \\b &= \frac{ n \cdot \sum{XY} - \sum{X} \cdot \sum{Y}}{n \cdot \sum{X^2} - \left(\sum{X}\right)^2} = \frac{ 10 \cdot 3108.8 - 45 \cdot 747.1 }{ 10 \cdot 285 - \left( 45 \right)^2} \approx -3.068\end{aligned}$

Step 4: Assemble the equation of a line

$\begin{aligned} y~&=~a ~+~ b \cdot x \\y~&=~88.518 ~-~ 3.068 \cdot x\end{aligned}$

The graph of the regression line is:

4 0

3 years ago

Miguel is a golfer, and he plays on the same course each week. The following table shows the probability distribution for his sc

aivan3 [116]

1) 4.55

2) Short hit

Step-by-step explanation:

The table containing the score and the relative probability of each score is:

Score 3 4 5 6 7

Probability 0.15 0.40 0.25 0.15 0.05

Here we call

X = Miguel's score on the Water Hole

The expected value of a certain variable X is given by:

$E(X)=\sum x_i p_i$

where

$x_i$ are all the possible values that the variable X can take

$p_i$ is the probability that $X=x_i$

Therefore in this problem, the expected value of MIguel's score is given by:

$E(X)=3\cdot 0.15 + 4\cdot 0.40 + 5\cdot 0.25 + 6\cdot 0.15 + 7\cdot 0.05=4.55$

In this problem, we call:

X = Miguel's score on the Water Hole

Here we have that:

- If the long hit is successfull, the expected value of X is

$E(X)=4.2$

- Instead, if the long hit fails, the expected value of X is

$E(X)=5.4$

Here we also know that the probability of a successfull long hit is

$p(L)=0.4$

Which means that the probabilty of an unsuccessfull long hit is

$p(L^c)=1-p(L)=1-0.4=0.6$

Therefore, the expected value of X if Miguel chooses the long hit approach is:

$E(X)=p(L)\cdot 4.2 + p(L^C)\cdot 5.4 = 0.4\cdot 4.2 + 0.6\cdot 5.4 =4.92$

In part 1) of the problem, we saw that the expected value for the short hit was instead

$E(X)=4.55$

Since the expected value for X is lower (=better) for the short hit approach, we can say that the short hit approach is better.

8 0

3 years ago

Help me ASAP!!

iVinArrow [24]

I believe you would add the 6m and 3m to make 9m then do the area equation which is a=1/2bh so 1/2 times 9 times 8 you would get 36 so that should be the answer 36

4 0

3 years ago