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charle [14.2K]
3 years ago
10

A teacher can spend as much as $95.00 to take students to a movie. A movie theater charges a $10.00 group fee and a $5.50 per ti

cket.
Which of the following inequalities represents the problem and could be solved to find the maximum number of students, x, who could attend the movie?

A: 5.5x + 10 < 95 (Wrong)
B: 5.5x + 10 ≥ 95
C: 5.5x + 10 ≤ 95
D: 5.5 + 10x ≤ 125
Mathematics
1 answer:
Pachacha [2.7K]3 years ago
3 0

Answer:

C

Step-by-step explanation:

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Write an equation in point slope form that passes through (-14, - 6) and the<br> slope is 2/3
s344n2d4d5 [400]

Answer:

y=2/3x - 6

Step-by-step explanation:

use the formula y=mx+b. the m of this equation is equal to the slope and you know that the slope is 2/3. the b of the equation is equal to the y-intercept so, to find b you simply look at the points given. the points are written in (x,y) form so if -14 is x, -6 must be y. remember to KEEP CHANGE CHANGE your signs so that plus sign becomes negative 6. Hope this helps!

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How many more games must Leon play in order to score at least 117 points .
kipiarov [429]

Answer:

how many does he have???

Step-by-step explanation:


8 0
3 years ago
X+8=21 I need helpppp
Mrac [35]
You add 13 and 8 it gets 21 so for X its 13
8 0
3 years ago
Read 2 more answers
A statistics textbook chapter contains 60 exercises, 6 of which are essay questions. A student is assigned 10 problems. (a) What
Scilla [17]

Answer:

a) P=0.3174

b) P=0.4232

c) P=0.2594

d) The shape of the hypergeometric, in this case, is like a binomial with mean np=1.

Step-by-step explanation:

The appropiate distribution to model this is the hypergeometric distribution:

P(X=x)=\frac{\binom{s}{x}\binom{N-s}{M-x}}{\binom{N}{M}}=\frac{\binom{6}{x}\binom{54}{10-x}}{\binom{60}{10}}

a) What is the probability that none of the questions are essay?

P(X=0)=\frac{\binom{6}{0}\binom{54}{10-0}}{\binom{60}{10}}\\\\P(X=0)=\frac{1*(54!/(10!*44!)}{60!/(10!*50!)} =\frac{2.3931*10^{10}}{7.5394*10^{10}} = 0.3174

b)  What is the probability that at least one is essay?

P(X=1)=\frac{\binom{6}{1}\binom{54}{9}}{\binom{60}{10}}\\\\P(X=1)=\frac{6*(54!/(9!*43!)}{60!/(10!*50!)} =\frac{3.1908*10^{10}}{7.5394*10^{10}} =0.4232

c) What is the probability that two or more are essay?

P(X\geq2)=1-(P(0)+P(1))=1-(0.3174+0.4232)=1-0.7406=0.2594

8 0
3 years ago
Neeeeed helpppppp please
MissTica

Answer:

Scale factor = 7

Step-by-step explanation:

Dilation with scale factor to map HEFG to DABC will be,

Scale factor = \frac{\text{Dimension of Image DABC}}{\text{Dimension of preimage HEFG}}

                    = \frac{\text{Length of CD}}{\text{Length of GH}}

Length of CD = Distance between two points C(0, -7) and D(-7, 0)

                       = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}

                       = \sqrt{(-7-0)^2+(0+7)^2}

                       = \sqrt{98}

                       = 7\sqrt{2}

Length of GH = Distance between G(0, -1) and H(-1, 0)

                       = \sqrt{(-1-0)^2+(0+1)^2}

                       = \sqrt{2}

Scale factor = \frac{7\sqrt{2} }{\sqrt{2} }

                    = 7

6 0
3 years ago
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