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tatuchka [14]
3 years ago
9

Can someone pls help me

Mathematics
2 answers:
svetoff [14.1K]3 years ago
4 0

18/24=0.75

?/80=.75

80x.75=?

?=60

5x+10=60

    -10  -10

5x=50

/5   /5

x=10

sub x in for 10

10x5=50+10=60

C.X=10

Anna007 [38]3 years ago
4 0
It’s Ccccccccccccccccccc
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Mr. Ortez told each student in his class to pick a folded piece of graph paper from a bowl. He instructed them to study the grap
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Answer:

a graph that connects 3 points connected by a straight line that passes through (5, 0)

Step-by-step explanation:

A graph is the representation of numerical data by one or more lines that give visibility to the relationship between the data. Understanding graphics today is an essential task, as they are very present in our daily lives, whether in newspapers, magazines, the internet, etc.

If a teacher asks each student to take a piece of graph paper containing a graph, study the graph and tell which graph represents graphs of equivalent proportions. The student who has a graph that connects 3 points connected by a straight line passing through (5, 0) is the student who should speak.

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William purchased 76.9 feet of fencing for his back yard. He realized after measuring, that he still needed an additional 25.7 f
Lostsunrise [7]
William needs 102.6 fencing in all because 76.9 plus 25.7 equals 102.6, hope this helps!
3 0
3 years ago
Weights and heights of turkeys tend to be correlated. For a population of turkeys at a farm, this correlation is found to be 0.6
LenaWriter [7]

Answer:

a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than <u>79.37 %</u> of them.

The  average height for turkeys at the 90th percentile for weight is 34.554

Of the turkeys at the 90th percentile for weight, roughly the percentage that  would  be taller than 28 inches 79.37%

Step-by-step explanation:

Given that:

For a population of turkeys at a farm, the correlation found between the weights and heights of turkeys is r = 0.64

the average weight in pounds \overline x = 17

the standard deviation of the weight in pounds S_x = 5

the average height in inches \overline y = 28

the standard deviation of the height in inches S_y = 8

Also, given that the weight and height both roughly follow the normal curve

For this study , the slope of the regression line can be expressed as :

\beta_1 = r \times ( \dfrac{S_y}{S_x})

\beta_1 = 0.64 \times ( \dfrac{8}{5})

\beta_1 = 0.64 \times 1.6

\beta_1 = 1.024

To the intercept of the regression line, we have the following equation

\beta_o = \overline y - \beta_1 \overline x

replacing the values:

\beta_o = 28 -(1.024)(17)

\beta_o = 28 -17.408

\beta_o = 10.592

However, the regression line needed for this study can be computed as:

\hat Y = \beta_o + \beta_1 X

\hat Y = 10.592 + 1.024 X

Recall that;

both the weight and height roughly follow the normal curve

As such, the weight related to 90th percentile can be determined as shown below.

Using the Excel Function at 90th percentile, which can be computed as:

(=Normsinv (0.90) ; we have the desired value of 1.28

∴

\dfrac{X - \overline x}{s_x } = 1.28

\dfrac{X - 17}{5} = 1.28

X - 17 = 6.4

X = 6.4 + 17

X = 23.4

The predicted height \hat Y = 10.592 + 1.024 X

where; X = 23.4

\hat Y = 10.592 + 1.024 (23.4)

\hat Y = 10.592 + 23.9616

\hat Y = 34.5536

Now; the probability of predicted height less than 34.5536 can be computed as:

P(Y < 34.5536) = P( \dfrac{Y - \overline y }{S_y} < \dfrac{34.5536-28}{8})

P(Y < 34.5536) = P(Z< \dfrac{6.5536}{8})

P(Y < 34.5536) = P(Z< 0.8192)

From the Z tables;

P(Y < 34.5536) =0.7937

Hence,  a turkey at the farm which weighs more than 90% of all the turkeys is predicted to be taller than <u>79.37 %</u> of them.

The  average height for turkeys at the 90th percentile for weight is :

\hat Y = 10.592 + 1.024 X

where; X = 23.4

\hat Y = 10.592 + 1.024 (23.4)

\hat Y = 10.592 + 23.962

\mathbf{\hat Y = 34.554}

Of the turkeys at the 90th percentile for weight, roughly what percent would you estimate to be taller than 28 inches?

i.e

P(Y >28) = 1 - P (Y< 28)

P(Y >28) = 1 - P( Z < \dfrac{28 - 34.554}{8})

P(Y >28) = 1 - P( Z < \dfrac{-6.554}{8})

P(Y >28) = 1 - P( Z < -0.8193)

From the Z tables,

P(Y >28) = 1 - 0.2063

\mathbf{P(Y >28) = 0.7937}

= 79.37%

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4 years ago
What is a slope intercept equation of the line that passes through the points (-6,-9) and (-11,-29)?
kiruha [24]
Y = 4x + 15 is slope intercept form
8 0
3 years ago
A maps key shows that every 5 inches on the map represents 200 miles of actual distance. Suppose the distance between two cities
ryzh [129]
Hello there, and thank you for posting your question here on brainly.

Short answer: 280 miles.

Why?

You can find this out by finding out how much 1 in is by dividing 200 by 5. (200 / 5 ==> 40) So 1 in = 40 miles. Now, we multiply that by 7, so we can find out how much 7 in would be. (7 * 4 ==> 280) 7 inches on the map represents 280 miles.

Hope this helped you! ♥
3 0
3 years ago
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