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4 years ago

13

The first pentagon is dilated to form the second pentagon. Drag and drop the answer to correctly complete the statement. IT IS 2

5 POINTS!!!

2 answers:

mafiozo [28]4 years ago

6 0

Its 1.25
Hope I helped C:

Yuki888 [10]4 years ago

6 0

Answer: The correct answer is 1.25.

Step-by-step explanation: We are given two similar figures that are regular pentagons with length of each side equal to 4 and 5.

We know that to find the scale factor of two similar figures, we need to find the ratio of length of the transformed figure to the length of the original figure.

Here, the first pentagon has side equal to 4 and the second pentagon has side equal to 5. So, the scale factor will be

$5:4=1.25.$

Thus, the scale factor is 1.25.

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A line passes through the point (-8, 3) and has a slope of 3/4. Write an equation in slope-intercept form for this line.

agasfer [191]

General formula to make slope-intercept form
y - y₁ = m(x - x₁)

From the question, we know that
m = 3/4
(x₁,y₁) = (-8,3)

Put the number to the formula
y - y₁ = m(x - x₁)
y - 3 = 3/4(x - (-8))
y - 3 = 3/4(x + 8)
y - 3 = (3/4)x + 6
y = (3/4)x + 9

6 0

4 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%

7 0

4 years ago

If y=2 when x=4,write down the value of y when x=6 for the following

lubasha [3.4K]

Nearly positive it’s 3

4 0

3 years ago

Day 3 please help i dont get it

jolli1 [7]

That's easy lol Its C,have a good day

8 0

3 years ago

Read 2 more answers

Each person in a group of twenty people at a hotel orders one meal chosen from oatmeal, eggs, or pancakes and one hot beverage c

Paraphin [41]

Answer:

$S = \{(Oatmeal, Coffee),(Eggs, Coffee),(Pancakes,Coffee),$ $(Oatmeal, Tea),(Eggs, Tea),(Pancakes,Tea)\}$

Step-by-step explanation:

Given

$Meal = \{Oatmeal, Eggs, Pancakes\}$

$Beverage = \{Coffee, Tea\}$

Required

Determine the sample space

The sample space is a list of combination of 1 meal and 1 beverage.

Because one individual can order only one of the meals and one of the beverages at any one time, then no two orders can be the same

The sample space (S) is as follows:

$S = \{(Oatmeal, Coffee),(Eggs, Coffee),(Pancakes,Coffee),$ $(Oatmeal, Tea),(Eggs, Tea),(Pancakes,Tea)\}$

8 0

3 years ago