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4 years ago

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5y + 4(-5 + 3y) = 1 – y solve step by step please

2 answers:

Yuri [45]4 years ago

8 0

Answer:

$\boxed {y = \frac{7}{6}}$

Step-by-step explanation:

Solve for the value of $y$ :

$5y + 4( -5 + 3y) = 1 - y$

-Use <u>Distributive Property</u>:

$5y + 4( -5 + 3y) = 1 - y$

$5y - 20 + 12y = 1 - y$

-Combine like terms:

$5y - 20 + 12y = 1 - y$

$17y - 20 = 1 - y$

-Take $-y$ and add it to $17y$ :

$17y + y - 20 = 1 - y + y$

$18y - 20 = 1$

-Add $20$ on both sides:

$18y - 20 + 20 = 1 + 20$

$18y = 21$

-Divide both sides by $18$ :

$\frac{18y}{18} = \frac{21}{18}$

$\boxed {y = \frac{7}{6}}$

Therefore, the value of $y$ is $\frac{7}{6}$ .

Studentka2010 [4]4 years ago

3 0

Step by step, meaning explaining the equation right and answer?

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Evaluate the expression 7x + y when x = 2 and y 6

sammy [17]

7(2) + 6
14 + 6
20
This is a answer = 20

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3 years ago

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In ΔBCD, the measure of ∠D=90°, DB = 96 feet, and CD = 14 feet. Find the measure of ∠B to the nearest tenth of a degree.

Reil [10]

Answer:

8.3

Step-by-step explanation:

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3 years ago

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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

Naina went on a trip. She covered a distance of 34 Kilometre 46 metre by car, 143 kilometre 92 metre by train, 5 kilometre 92 me

GaryK [48]

Answer:

185590 m

or 185 km 590 meters

Step-by-step explanation:

to calculate total distance covered, convert km to meter and adding the numbers together

1km = 1000m

34 Kilometre 46 metre = (34 x 1000) + 46 = 34046m

143 kilometre 92 metre = ( 143 x 1000) + 92 = 143,092m

5 kilometre 92 metre = (5 x 1000) + 92 = 5092

2 kilometre 60 metre = (2 x 1000) + 60 = 2060

1.3 km = 1.3 x 1000 = 1300

7 0

3 years ago

1^2 + 2^2 + 3^2 +.... + 28^2 + 29^2 + 30^2 = ......<br> Find the value <br> Thank you!

liberstina [14]

Step-by-step explanation:

The sum of consecutive squares is

$1^2+2^2+\dots+n^2=\frac16n(n+1)(2n+1)$

Therefore

$1^2+2^2+\dots+30^2=\frac16(30)(31+1)(2\cdot30+1) = 9455$

8 0

3 years ago