Will's Pizza can make 12 pizzas in 45 minutes. How many pizzas can they make in 4 hours?

Mathematics

2 answers:

irina1246 [14]3 years ago

8 0

Answer:

64 pizzas

Step-by-step explanation:

4 hours is 240 minutes

12 x (240 / 45) = 64

babymother [125]3 years ago

8 0

Answer: 64

Step-by-step explanation:

Alright. We are trying to find out how many pizza's they can make in every 15 mins so we can find out each hour. 45 divided by 3 is 15, and 12 divided by 3 is 4, so they can make 4 pizzas in 15 mins and multiply that by 4 is 16 pizzas in 1 hour. We are looking for 4 hours though so we just multiply 16 by 4 and we get 64.

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Write first three terms of AP with a = -1 and d = 1/2

marta [7]

Answer:

first three terms are

-1, -1/2 & 0

Step-by-step explanation:

here,

a=-1

and d=1/2

t1 = -1

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t3= a+(n-1)d= -1 +(3-1)*(1/2) =0

6 0

3 years ago

A national college researcher reported that 65% of students who graduated from high school in 2012 enrolled in college. Twenty n

Usimov [2.4K]

Answer:

a) The probability that exactly 17 of them enroll in college is 0.116.

b) The probability that more than 14 enroll in college is 0.995.

c) The probability that fewer than 11 enroll in college is 0.001.

d) It would be be unusual if more than 24 of them enroll in college since the probability is 0.009.

Step-by-step explanation:

We can model this with a binomial distribution, with n=29 and p=0.65.

The probability that k students from the sample who graduated from high school in 2012 enrolled in college is:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{29}{k} 0.65^{k} 0.35^{29-k}\\\\\\$

a) The probability that exactly 17 of them enroll in college is:

$P(x=17) = \dbinom{29}{17} p^{17}(1-p)^{12}=51895935*0.0007*0=0.116\\\\\\$

b) The probability that more than 14 of them enroll in college is:

$P(X>14)=\sum_{15}^{29} P(X=k_i)=1-\sum_{0}^{14} P(X=k_i)\\\\\\P(x=0)=0\\\\P(x=1)=0\\\\P(x=2)=0\\\\P(x=3)=0\\\\P(x=4)=0\\\\P(x=5)=0\\\\P(x=6)=0\\\\P(x=7)=0\\\\P(x=8)=0\\\\P(x=9)=0\\\\P(x=10)=0.001\\\\P(x=11)=0.002\\\\P(x=12)=0.005\\\\P(x=13)=0.013\\\\P(x=14)=0.027\\\\\\P(X>14)=1-0.005=0.995$

c) Using the probabilities calculated in the point b, we have:

$P(X$

d) The probabilities that more than 24 enroll in college is:

$P(X>24)=\sum_{25}^{29}P(X=k_i)\\\\\\ P(x=25) = \dbinom{29}{25} p^{25}(1-p)^{4}=23751*0*0.015=0.007\\\\\\P(x=26) = \dbinom{29}{26} p^{26}(1-p)^{3}=3654*0*0.043=0.002\\\\\\P(x=27) = \dbinom{29}{27} p^{27}(1-p)^{2}=406*0*0.123=0\\\\\\P(x=28) = \dbinom{29}{28} p^{28}(1-p)^{1}=29*0*0.35=0\\\\\\P(x=29) = \dbinom{29}{29} p^{29}(1-p)^{0}=1*0*1=0\\\\\\\\P(X>24)=0.007+0.002+0+0+0=0.009$