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Illusion [34]
2 years ago
14

The archway of the main entrance of a university is modeled by the quadratic equation y = -x2 + 6x. The university is hanging a

banner at the main entrance at an angle defined by the equation 4y = 21 − x. At what points should the banner be attached to the archway?
A(1, 5.5) and (5.25, 6.56)
B(1, 5) and (5.25, 3.94)
C(1.5, 4.87) and (3.5, 4.37)
D(1.5, 5.62) and (3.5, 6.12)
EThere is no real solution.
Mathematics
2 answers:
katrin [286]2 years ago
5 0

Basically look for the intersection points of these two functions by setting them equal to each other.

-x^2 + 6x = 21/4 - x/4

-x^2 + (25/4)x - 21/4 = 0

x = 1 and x = 21/4


Plug these points back to any of the two original equations, you will get

(1,5) and (21/4, 63/16)


denpristay [2]2 years ago
4 0

1.5 ........ 5.25..... 3.94

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Reyna has 5 coins worth 10 cents each and 4 coins
Zolol [24]

Answer:

The Probability found is:

P =  \frac{13}{18}

Step-by-step explanation:

Let x be the 10 cents coin.

Let y be the 25 cents coin.

We have to find all the possible outcomes

1) First coin = 10 cents, Second coin = 10 cents , so

(x,x) = 20

2) First coin = 10 cents, Second coin = 25 cents , so

(x,y) = 35

3) First coin = 25 cents, Second coin = 10 cents , so

(y,x) = 35

4) First coin = 25 cents, Second coin = 25 cents , so

(y,y) = 50

Find the probability of each outcome:

P(x,x) =  \frac{5}{9}\cdot\frac{4}{8}=\frac{20}{72}

P(x,y) =  \frac{5}{9}\cdot\frac{4}{8}=\frac{20}{72}

P(y,x) =  \frac{5}{9}\cdot\frac{4}{8}=\frac{20}{72}

P(y,y) = \frac{4}{9}\cdot\frac{3}{8}=\frac{12}{72}

Add all the probabilities where sum is at least 35 i.e P(x,y) , P(y,x) , P(y,y)

P(x,y) + P(y,x) + P(y,y) = \frac{20}{72}+\frac{20}{72}+\frac{12}{72} = \frac{52}{72}=\frac{13}{18}\\

6 0
2 years ago
Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.
Ghella [55]

Answer:  The required inequality is x+\dfrac{1}{4}x+\dfrac{3}{2}x>22. and its solution is x>8.

Step-by-step explanation:  Given that Mustafa, Heloise, and Gia have written more than a combined total of 22 articles for the school newspaper.

Also, Heloise has written \frac{1}{4} as many articles as Mustafa has and Gia has written \frac{3}{2} as many articles as Mustafa has.

We are to write an inequality to determine the number of articles, m, Mustafa could have written for the school newspaper.  Also, to solve the inequality.

Since m denotes the number of articles that Mustafa could have written. Then, according to the given information, we have

x+\dfrac{1}{4}x+\dfrac{3}{2}x>22.

And the solution of the above inequality is as follows :

x+\dfrac{1}{4}x+\dfrac{3}{2}x>22\\\\\\\Rightarrow \dfrac{4x+x+6x}{4}>22\\\\\\\Rightarrow 11x>88\\\\\Rightarrow x>\dfrac{88}{11}\\\\\Rightarrow x>8.

Thus, the required inequality is x+\dfrac{1}{4}x+\dfrac{3}{2}x>22. and its solution is x>8.

4 0
2 years ago
Read 2 more answers
In general, after a group makes a reservation at a restaurant, they do not show up 20% of the time. If a restaurant has 8 reserv
sashaice [31]

Answer:

P(more than 2 don't show up)=0.2031

Step-by-step explanation:

Multiplication rule for independent events:

If two events A and B are independent then the multiplication rule state that:

P(A and B)= P(A) x P(B)

Find the probability that more than two do not show up:

The probability that the reservations at the restaurant do not show up is 0.20 (=20%) and there were 8 reservations.

P(more than 2 don't show up)= 1- P(less than or equal to 2 don't show up)

= 1-( P(none show up) +P(one don't show up)+P(two don't show up)

=1- ((0.8⁸) +₈ C₁ (0.2¹)x(0.8⁷)  +₈ C₂ (0.2²)x(0.8⁶)

=1- (0.1678 +0.3355+0.2936)= 1-0.7969

P(more than 2 don't show up)=0.2031

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2 years ago
Sumy is working in geometry class and is given figure ABCD in the coordinate plane to reflect. The coordinates of point D are (a
Makovka662 [10]

Answer:

(b, a)

Step-by-step explanation:

Example if D was (3, 2) and reflects across y = x, D' will be (2, 3)

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A polling agency reported that 66 percent of adults living in the United States were satisfied with their health care plans. The
Tpy6a [65]

Answer:

A) The probability is 0.95 that the percent of adults living in the United States who are satisfied with their health care plans is between 63.6% and 68.4%.

Step-by-step explanation:

A polling agency reported that 66 percent of adults living in the United States were satisfied with their health care plans. The estimate was taken from a random sample of 1,542 adults living in the United States, and the 95 percent confidence interval for the population proportion was calculated as (0.636, 0.684).

This means that we are 95% sure that the true proportion of adults living in the United States who were satisfied with their health care plans is between 0.636 and 0.684.

So the correct answer is:

A) The probability is 0.95 that the percent of adults living in the United States who are satisfied with their health care plans is between 63.6% and 68.4%.

7 0
2 years ago
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