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katen-ka-za [31]
2 years ago
11

Izaak is a big baseball fan. He is comparing the cost of two options for purchasing tickets. Let g represent the number of baseb

all games Izaak plans to attend. Let C represent the cost. Single- game tickets: C = 8g Season tickets: C = 1.50g+195
Under what circumstances is the cost of single game tickets less than the cost of season tickets ?
Mathematics
1 answer:
Schach [20]2 years ago
6 0

The single game ticket only costs 8g per ticket. The Season tickets costs 1.19 (I had added them). The 8 is lesser than 1.19. So the Season tickets costs more than the Single Game tickets (I hope this helped!).

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At a pizzeria, the ratio of pizzas topped with meat to pizzas topped with vegetables was 4:3 on Saturday. The pizzeria made a to
vivado [14]

Answer:

Option B is correct

36 pizzas were topped with vegetables

Step-by-step explanation:

Let x be the number.

As per the statement:

At a pizzeria, the ratio of pizzas topped with meat to pizzas topped with vegetables was 4:3 on Saturday.

The ratio of pizza topped with meat to pizzas topped with vegetables = 4 : 3

then;

the number of pizza topped with meat = 4x

and

the number pizza topped with vegetables  = 3x

It is also given that the pizzeria made a total of 84 pizzas that day.

\text{Total Pizzas} = \text{Number of pizzas topped with meat}+ \text{Number of pizzas topped with vegetables}

Substitute the given values we get;

84 = 4x+3x

Combine like terms;

84= 7x

Divide both sides by 7 we get;

12 = x

The pizza were topped with vegetables = 3x = 3(12) = 36

therefore, 36 pizzas were topped with vegetables


6 0
2 years ago
Read 2 more answers
The weight of Jacob’s backpack is made up of the weight of the contents of the backpack as well as the weight of the backpack it
Dafna11 [192]
If the backpack (b) contains only textbooks (t) and notebooks (n), the total weight of the backpack is:

b+t+n

but notebooks are 4 pounds, so actually:

b+t+4

also, the backpack weights 2 pounds, so the total weight is:

2+t+4 =t+6
 now, "Seventy percent of <span> the total weight is textbooks. ", this means that


t=70%(t+6) - this equation can be used to determine the weight of the textbooks!
</span> 




7 0
2 years ago
Read 2 more answers
An automated egg carton loader has a 1% probability of cracking an egg, and a customer will complain if more than one egg per do
vaieri [72.5K]

Answer:

a) Binomial distribution B(n=12,p=0.01)

b) P=0.007

c) P=0.999924

d) P=0.366

Step-by-step explanation:

a) The distribution of cracked eggs per dozen should be a binomial distribution B(12,0.01), as it can model 12 independent events.

b) To calculate the probability of having a carton of dozen eggs with more than one cracked egg, we will first calculate the probabilities of having zero or one cracked egg.

P(k=0)=\binom{12}{0}p^0(1-p)^{12}=1*1*0.99^{12}=1*0.886=0.886\\\\P(k=1)=\binom{12}{1}p^1(1-p)^{11}=12*0.01*0.99^{11}=12*0.01*0.895=0.107

Then,

P(k>1)=1-(P(k=0)+P(k=1))=1-(0.886+0.107)=1-0.993=0.007

c) In this case, the distribution is B(1200,0.01)

P(k=0)=\binom{1200}{0}p^0(1-p)^{12}=1*1*0.99^{1200}=1* 0.000006 = 0.000006 \\\\ P(k=1)=\binom{1200}{1}p^1(1-p)^{1199}=1200*0.01*0.99^{1199}=1200*0.01* 0.000006 \\\\P(k=1)= 0.00007\\\\\\P(k\leq1)=0.000006+0.000070=0.000076\\\\\\P(k>1)=1-P(k\leq 1)=1-0.000076=0.999924

d) In this case, the distribution is B(100,0.01)

We can calculate this probability as the probability of having 0 cracked eggs in a batch of 100 eggs.

P(k=0)=\binom{100}{0}p^0(1-p)^{100}=0.99^{100}=0.366

5 0
2 years ago
gavin is making cranberry lemonade. he wants to make a total of 6 cups. he needs to use 3 times as much lemonade as cranberry ju
Inessa [10]

Answer:

3l= c

Step-by-step explanation:

as use 3 cup of cranberry juice, lemonade will be used 1 cup.

4 0
2 years ago
From 2000-2003, students were tested by the state in four major subject areas of math, science, English and social studies. The
Verizon [17]

No way to tell . . . . . we can't see the chart below.
It must be WAY down there where the sun don't shine.
8 0
2 years ago
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