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amid [387]
3 years ago
13

Need help will give 5 stars.

Mathematics
2 answers:
olga_2 [115]3 years ago
7 0

Answer:

t=0.64

Step-by-step explanation:

h = -16t^2 +4t +4

We want h =0 since it is hitting the ground

0 = -16t^2 +4t +4

Using the quadratic formula

a = -16  b = 4  c=4

-b ± sqrt( b^2 -4ac)

----------------------------

         2a

-4 ± sqrt( 4^2 -4(-16)4)

----------------------------

         2(-16)

-4 ± sqrt( 16+ 256)

----------------------------

         -32

-4 ± sqrt( 272)

----------------------------

         -32

-4 ± sqrt( 16*17)

----------------------------

         -32

-4 ± sqrt( 16) sqrt(17)

----------------------------

         -32

-4 ± 4 sqrt(17)

----------------------------

         -32

Divide by -4

1 ±  sqrt(17)

----------------------------

         8

To the nearest hundredth

t=-0.39

t=0.64

Since time cannot be negative

t=0.64

ICE Princess25 [194]3 years ago
5 0

Answer:

0.64  

Step-by-step explanation:

0 = -16t^2 + 4t + 4

-4(4t^2 - t -1) = 0

t = [-(-1) +/- sqrt (1 - 4*4*-1)] / 8)

t = 0.64, -0.39

answer is 0.64

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Past records indicate that the probability of online retail orders that turn out to be fraudulent is 0.08. Suppose that, on a gi
Sunny_sXe [5.5K]

Answer:

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

Step-by-step explanation:

We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.08.

The probability of k online retail orders that turn out to be fraudulent in the sample is:

P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.08^k\cdot0.92^{20-k}

We have to calculate the probability that 2 or more online retail orders that turn out to be fraudulent. This can be calculated as:

P(x\geq2)=1-[P(x=0)+P(x=1)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.08^{0}\cdot0.92^{20}=1\cdot1\cdot0.189=0.189\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.08^{1}\cdot0.92^{19}=20\cdot0.08\cdot0.205=0.328\\\\\\\\P(x\geq2)=1-[0.189+0.328]\\\\P(x\geq2)=1-0.517=0.483

The probability that there are 2 or more fraudulent online retail orders in the sample is 0.483.

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3 years ago
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