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alex41 [277]
3 years ago
13

For a quarterback, the probability of throwing a completed pass is 60%, and the probability of gaining more than 10 yards on a r

unning play is 25%. What is the probability of gaining more than 10 yards on a running play and then throwing a completed pass? Express your answer as a decimal rounded to the nearest hundredth. What is the probability of throwing 3 completed passes in a row? Express your answer as a decimal rounded to the nearest hundredth.
Mathematics
1 answer:
never [62]3 years ago
3 0
The probability of gaining more than 10 yards then throwing a completed pass is 15%.

To find the chance of both events happening, you multiply the percents.
0.25 x 0.6 = 0.15 or 15%

The chance of throwing 3 completed passes would be:
0.6 x 0.6 x 0.6 = 0.216 or 21.6%
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Michael sold 23 of his CDs and purchased 16 more.
aleksley [76]
25-16=9
9+23=32
He had 32 CDs to begin with
6 0
3 years ago
What should I buy? A study conducted by a research group in a recent year reported that of cell phone owners used their phones i
Llana [10]

Answer:

The probability that seven or more of them used their phones for guidance on purchasing decisions is 0.7886.

<em />

Step-by-step explanation:

<em>The question is incomplete:</em>

<em>What should I buy? A study conducted by a research group in a recent year reported that 57% of cell phone owners used their phones inside a store for guidance on purchasing decisions. A sample of 14 cell phone owners is studied. Round the answers to at least four decimal places. What is the probability that seven or more of them used their phones for guidance on purchasing decisions? </em>

We can model this as a binomial random variable, with p=0.57 and n=14.

P(x=k)=\dbinom{n}{k} p^{k}q^{n-k}

a) We have to calculate the probability that seven or more of them used their phones for guidance on purchasing decisions:

P(x\geq7)=\sum_{k=7}^{14}P(x=k)\\\\\\

P(x=7)=\dbinom{14}{7} p^{7}q^{7}=3432*0.0195*0.0027=0.1824\\\\\\P(x=8) = \dbinom{14}{8} p^{8}q^{6}=3003*0.0111*0.0063=0.2115\\\\\\P(x=9) = \dbinom{14}{9} p^{9}q^{5}=2002*0.0064*0.0147=0.1869\\\\\\P(x=10) = \dbinom{14}{10} p^{10}q^{4}=1001*0.0036*0.0342=0.1239\\\\\\P(x=11) = \dbinom{14}{11} p^{11}q^{3}=364*0.0021*0.0795=0.0597\\\\\\P(x=12) = \dbinom{14}{12} p^{12}q^{2}=91*0.0012*0.1849=0.0198\\\\\\P(x=13) = \dbinom{14}{13} p^{13}q^{1}=14*0.0007*0.43=0.004\\\\\\

P(x=14) = \dbinom{14}{14} p^{14}q^{0}=1*0.0004*1=0.0004\\\\\\

P(x\geq7)=0.1824+0.2115+0.1869+0.1239+0.0597+0.0198+0.004+0.0004\\\\P(x\geq7)=0.7886

7 0
3 years ago
Plzzzzz help!!!!! I will give brainliest
Vitek1552 [10]

Answer:

<h2>10.88inches</h2>

Step-by-step explanation:

3 0
3 years ago
a 10 gram sample of a substance that's used to detect explosives has k-value of 0.1353, find the substance half life in days, Ro
IrinaK [193]
We find the value of N₀ since we are provided with initial conditions.
The condition is that, at time t = 0, the amount of substance contains originally 10 grams.
We substitute:
10 = N₀ (e^(-0.1356)*0)
10 = N₀ (e^0)
N₀ = 10

When the substance is in half-life (meaning, the half of the original amount), it contains 5 grams. We solve t in this case.
5 = 10 e^(-0.1356*t)
0.5 = e^(-0.1356*t)
Multiply natural logarithms on both sides to bring down t.
ln(0.5) = -0.1356*t
Hence,
t = -(ln(0.5))/0.1356
t ≈ 5.11 days (ANSWER)
4 0
3 years ago
Read 2 more answers
Translate this sentence into an equation.
mr Goodwill [35]

Answer:

r - 12 = 74

Step-by-step explanation:

The difference between his height with 12 would signify subtraction with 12, and that difference is 74. Meaning, his height, minus 12, would equal 74.

6 0
3 years ago
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