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Shkiper50
2 months ago
8

If NO = 17 and NP = 5x -6, find the value of x. Then find NO and OP.

Mathematics
1 answer:
Zina [12.3K]2 months ago
7 0
Assuming that NO equals NP, we can set up the equation 17 = 5x - 6. To find x, we add 6 to both sides, resulting in 23 = 5x. I'm not sure if the answer should be in decimal form, but dividing 23 by 5 gives 4.6. Therefore, x equals 4.6. Substituting this back, we find NO = 17 and NP = 17, which I believe is the solution to the problem. Please let me know if this is correct.
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Determine if each scenario is either a permutation or combination. Do NOT solve these scenarios. a) An art gallery displays 125
PIT_PIT [12445]

Answer:

a) Permutaciones

b) combinaciones

Step-by-step explanation:

a)

Dado que el orden de las 15 obras de arte más importantes es relevante, comenzando por la más popular y luego las que ocupan los lugares 2, 3, 4, y así sucesivamente. Cuando nos ocupamos del "orden de colocación", el tema se refiere a permutaciones.

b)

De un total de 320 obras, se deben "seleccionar" 125 para ser exhibidas, el proceso de selección implica combinaciones en las que el orden de la selección no importa.

3 0
2 months ago
A pineapple is 7 times as heavy as an orange. the pineapple also weighs 870 grams more than the orange. what is the total weight
zzz [12365]
Let's convert each statement into mathematical terms.

Pineapple = p

Orange = x

is indicates equality

 

Pineapple has a weight that is 7 times that of an orange:  p = 7x

Pineapple is 870 grams heavier than the orange. p = 870 + x

 

Insert  p = 7x  into the second equation:

7x = 870 + x

Subtract  x from both sides

6x = 870

Divide by 6

x = 145

 

The weight of the orange is 145 grams.

The weight of the pineapple is 7 times that of the orange = 7(145) = 1015 grams.

4 0
1 month ago
When the sun is at a certain angle in the sky, a 50-foot building will cast a -foot shadow. What is the length of the shadow in
Svet_ta [12734]

Answer:

0.40 feet

Step-by-step explanation:

For the first scenario, a 50-foot building casts a shadow of 1 foot. Let the angle of elevation of the sun from the shadow be denoted as θ.

Then:

Tan θ = \frac{opposite}{adjacent}

Tan θ = \frac{50}{1}

Tan θ = 50

⇒ θ = Tan^{-1} 50

      = 88.8542

      = 88.85^{o}

The elevation angle is roughly 88.85^{o}.

For a 20-foot pole,

Tan θ = \frac{opposite}{adjacent}

Tan 88.85^{o} = \frac{20}{x}

x = \frac{20}{Tan 88.85^{o} }

 = 0.4015

 = 0.40 feet

The length of the pole's shadow is 0.40 feet.

4 0
1 month ago
How many weeks of flea treatment would Jim’s dog get from one package if each treatment only lasted 3 weeks In a normal year whe
Inessa [12570]

Solution

The number of weeks in a year is 52 weeks

In a typical year, where each flea treatment lasts 4 weeks, Jim's dog would receive

\frac{52}{4} =13 times

Conversely, during especially bad flea years, when treatments only last for 3 weeks, Jim's dog would need

So in a year, Jim's dog would need 17- 13 =4 extra treatments.

4 treatments at 3 weeks each total 4×3 =12 weeks of additional treatment

4 0
1 month ago
A random sample of 20 individuals who graduated from college five years ago were asked to report the total amount of debt (in $)
AnnZ [12381]

Response:

a. As student debt rises, current investment diminishes.

b. Y= 68778.2406 - 1.9112X

For each dollar increase in college debt, the average current investments decrease by 1.9112 dollars.

c. A substantial linear correlation exists between college debt and current investment as the P-value falls below 0.1.

d. Y= $59222.2406

e. R²= 0.9818

Step-by-step breakdown:

Hello!

Data has been gathered on a random sample of 20 individuals who completed their college education five years ago. The variables under consideration are:

Y: Current investment by an individual who graduated from college five years prior.

X: Total debt of an individual upon graduating five years ago.

a)

To explore the relationship between debt and investment, creating a scatterplot with the sample data is ideal.

The scatterplot demonstrates a negative correlation, indicating that as these individuals' debt increases, their current investments decrease.

Therefore, the statement that accurately describes this is: As college debt rises, current investment decreases.

b)

The population regression equation is Y= α + βX +Ei

To develop this equation, estimates for alpha and beta are required:

a= Y[bar] -bX[bar]

a= 44248.55 - (-1.91)*12829.70

a= 68778.2406

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} }

b=\frac{9014653088-\frac{(256594)(884971)}{20} }{4515520748-\frac{(256594)^2}{20} }

b= -1.9112

∑X= 256594

∑X²= 4515520748

∑Y= 884971

∑Y²= 43710429303

∑XY= 9014653088

n= 20

Averages:

Y[bar]= ∑Y/n= 884971/20= 44248.55

X[bar]= ∑X/n= 256594/20= 12829.70

The estimated regression equation becomes:

Y= 68778.2406 - 1.9112X

For every dollar increase in college debt, the average current investments drop by 1.9112 dollars.

c)

To evaluate if there's a linear regression between these variables, the following null hypotheses are formulated:

H₀: β = 0

H₁: β ≠ 0

α: 0.01

Testing can be performed utilizing either a Student t-test or Snedecor's F (ANOVA)

Using t=  b - β  =  -1.91 - 0  = -31.83

                 Sb         0.06

The critical area and P-value for this test is two-tailed. The P-value equals: 0.0001

Since this P-value is underneath the significance level, we reject the null hypothesis.

In the case of ANOVA, the rejection area is also one-tailed to the right, corresponding to the P-value.

The P-value remains: 0.0001

Using this method, we similarly reject the null hypothesis.F= \frac{MSTr}{MSEr}= \frac{4472537017.96}{4400485.72} =1016.37

In conclusion, at a significance level of 1%, there exists a linear relationship linking current investment to college debt.

The accurate statement is:

There exists a significant linear association between college debt and current investment since the P-value is less than 0.1.

d)

To forecast the value of Y when X is set, it is essential to substitute X in the estimated regression equation.

Y/$5000

Y= 68778.2406 - 1.9112*5000

Y= $59222.2406

The anticipated investment for someone with a college debt of $5000 is $59222.2406.

e)

To determine the proportion of variation in the dependent variable that the independent variable accounts for, the coefficient of determination R² must be calculated.

R²= 0.9818

R^2= \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{sumY^2-\frac{(sumY)^2}{n} }

R^2= \frac{-1.9112^2[4515520748-\frac{(256594)^2}{20} ]}{43710429303-\frac{(884971)^2}{20} }

This indicates that 98.18% of the variability in current investments relates to college graduation debt within the projected regression model: Y= 68778.2406 - 1.9112X

I trust this is beneficial!

5 0
1 month ago
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