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olganol [36]
3 years ago
13

A football stadium holds 52,000 fans. A college student is doing research and determines that on any given game day, the home te

am has five times as many fans as the visiting team. In order to help the student in his research, he represents the number of home team tickets as H and the visiting team’s tickets as V.
Mathematics
1 answer:
allochka39001 [22]3 years ago
7 0
If H represents the tome teams tickets and V represents the visiting teams then it would be v+h(5)=52,000. 
V= 8,667 tickets 
H= 43,333 tickets
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Factor completely. X^2-14x+48<br> From the factor quadratic binomial quiz
ivanzaharov [21]

Answer: (x−8)(x−6)

Step-by-step explanation:  Factor x2−14x+48

using the AC method.

Hope this helped

:D

3 0
3 years ago
A manufacturer of personal computers sets tests competing brands and finds that the amounts of energy they require are normally
jeyben [28]

Answer:

b. [278.90, 288.55]

See the explanation below.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the amounts of energy of a population, and for this case we know the distribution for X is given by:

X \sim N(285,9.1)  

Where \mu=285 and \sigma=9.1

LOWER LIMIT

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.75   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.25 of the area on the left and 0.75 of the area on the right it's z=-0.674. On this case P(Z<-0.674)=0.25 and P(z>-0.674)=0.75

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=-0.674

And if we solve for a we got

a=285 -0.674*9.1=278.90

So the value of height that separates the bottom 25% of data from the top 75% is 278.90.  

UPPER LIMIT

For this part we want to find a value a, such that we satisfy this condition:

P(X>a)=0.3   (a)

P(X   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.7 of the area on the left and 0.3 of the area on the right it's z=0.524. On this case P(Z<0.524)=0.7 and P(z>0.524)=0.3

If we use condition (b) from previous we have this:

P(X  

P(z

But we know which value of z satisfy the previous equation so then we can do this:

z=0.524

And if we solve for a we got

a=285 +0.524*9.1=289.78

So the value of height that separates the bottom 70% of data from the top 30% is 289.78.  

With the procedure we got for the limits [278.9 , 289.78]

And since any of the options are on the list we take th most similar for this case:

b. [278.90, 288.55]

4 0
4 years ago
Which expressions are equivalent? <br> need done asap .
PIT_PIT [208]

Answer:

3x - 7y and -7y + 3x

6 0
2 years ago
Which of there polynomials could have (x-2) as a factor?
sertanlavr [38]

Answer:

x-2 is a factor of following polynomials:

  • b(x) = 3x² + 15x - 42
  • d(x) = 3x³ - 2x² - 15x + 14
  • e(x) = 8x⁴ - 41x³ - 18x² + 101x + 70

Step-by-step explanation:

In order to check that which polynomials might have x-2 as a factor, we will put

x-2 = 0 => x = 2 in each polynomial, if the value of polynomial on 2 is zero then x-2 is a factor otherwise not.

So,

a(x) = 6x² - 7x - 5

Putting x = 2

a(2) = 6(2)^2-7(2)-5\\= 6(4)-14-5\\=24-14-5\\=24-19\\=5 \neq 0

b(x) = 3x² + 15x - 42

b(2) = 3(2)^2 + 15(2) - 42\\=3(4)+30-42\\=12+30-42\\=42-42 = 0

c(x) = 2x³ + 13x² + 16x + 5

c(2) = 2(2)^3 + 13(2)^2 + 16(2) + 5\\=2(8)+13(4)+32+5\\=16+52+32+5\\=105 \neq 0

d(x) = 3x³ - 2x² - 15x + 14

d(2) = 3(2)^3 - 2(2)^2 - 15(2) + 14\\= 3(8)-2(4)-30+14\\=24-8-30+14\\=38-38 = 0

e(x) = 8x⁴ - 41x³ - 18x² + 101x + 70

e(2) = 8(2)^4 - 41(2)^3 - 18(2)^2 + 101(2) + 70\\= 8(16)-41(8)-18(4)+202+70\\=128-328-72+202+70\\=0

f(2) = (2)^4 + 5(2)^3 - 27(2)^2 - 101(2) - 70\\=16+5(8)-27(4)-202-70\\=16+40-108-202-70\\=-324 \neq 0

Hence,

x-2 is a factor of following polynomials:

  • b(x) = 3x² + 15x - 42
  • d(x) = 3x³ - 2x² - 15x + 14
  • e(x) = 8x⁴ - 41x³ - 18x² + 101x + 70
8 0
3 years ago
Help these are hard<br> Pls quick
ratelena [41]
19.2 × 21 = 403.2
He drove a total of 403.2 miles to and from work in 21 days.
4 0
4 years ago
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