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Llana [10]
3 years ago
6

GRADUATING THIS WEEK NEED THIS ALL ANSWERED

Mathematics
2 answers:
Elza [17]3 years ago
6 0

Answer:

3 hours

Step-by-step explanation:

We have the amount of miles that Tanika passed, 189 miles. We also have the speed at which Tanika was driving, 63 miles per hour. Basically, Tanika was passing 63 miles on every one hour of driving. In order to find out how much time was Tanika driving with that speed to pass those miles, we just simply need to divide the miles passed with the speed of driving:

189 / 63 = 3

So we have a result of 3, thus Tanika needed three hours with a speed of 63 miles per hour to pass 189 miles.

aalyn [17]3 years ago
5 0

Question 1:

For this case we can raise a rule of three:

63 miles ---------------> 1 hour

189 miles -------------> x

Where "x" represents the number of hours it takes Tanika to travel 189 miles.

x = \frac {189 * 1} {63}\\x = 3

So, Tanika takes 3 hours to travel 189 miles

Answer:

Three hours

Question 2:

For this case we must solve the following equations:

A) -2x

Dividing between -2 on both sides of the inequality:

x

B) 5t + 7

We subtract 7 on both sides of the inequality:

5t

Dividing between 5 on both sides of the inequality:

t

C) 12s-17 \geq2s + 33

We subtract 2s on both sides of the inequality:

12s-2s-17 \geq33\\10s-17 \geq33

We are 17 on both sides of the inequality:

10s \geq33 + 17\\10s \geq50

We divide between 10 on both sides of the inequality:

s \geq \frac {50} {10}\\s \geq5

D) -8m> -24

Dividing between -8 on both sides of the inequality:

m> \frac {-24} {- 8}\\m> 3

E) 8n + 7> 4n + 35

Subtracting 4n on both sides of the inequality:

8n-4n> 35-7\\4n> 28

Dividing between 4 on both sides of the inequality:

n> \frac {28} {4}\\n> 7

Answer:

x  7

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Answer:  The required answers are

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Step-by-step explanation:  Given that we toss a fair coin 10 times and X denote the number of heads.

We are to find

(a) the probability that X=5

(b) the probability that X greater or equal than 5

(c) the minimum value of a such that P(X ≤ a) > 0.8.

We know that the probability of getting r heads out of n tosses in a toss of coin is given by the formula of binomial distribution as follows :

P(X=r)=^nC_r\left(\dfrac{1}{2}\right)^r\left(\dfrac{1}{2}\right)^{n-r}.

(a) The probability of getting 5 heads is given by

P(X=5)\\\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}\\\\\\=\dfrac{10!}{5!(10-5)!}\dfrac{1}{2^{10}}\\\\\\=0.24609\\\\\sim0.25.

(b) The probability of getting 5 or more than 5 heads is

P(X\geq 5)\\\\=P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\\\\=^{10}C_5\left(\dfrac{1}{2}\right)^5\left(\dfrac{1}{2}\right)^{10-5}+^{10}C_6\left(\dfrac{1}{2}\right)^6\left(\dfrac{1}{2}\right)^{10-6}+^{10}C_7\left(\dfrac{1}{2}\right)^7\left(\dfrac{1}{2}\right)^{10-7}+^{10}C_8\left(\dfrac{1}{2}\right)^8\left(\dfrac{1}{2}\right)^{10-8}+^{10}C_9\left(\dfrac{1}{2}\right)^9\left(\dfrac{1}{2}\right)^{10-9}+^{10}C_{10}\left(\dfrac{1}{2}\right)^{10}\left(\dfrac{1}{2}\right)^{10-10}\\\\\\=0.24609+0.20507+0.11718+0.04394+0.0097+0.00097\\\\=0.62295\\\\\sim 0.62.

(c) Proceeding as in parts (a) and (b), we see that

if a = 10, then

P(X\leq 0)=0.00097,\\\\P(X\leq 1)=0.01067,\\\\P(X\leq 2)=0.05461,\\\\P(X\leq 3)=0.17179,\\\\P(X\leq 4)=0.37686,\\\\P(X\leq 5)=0.62295,\\\\P(X\leq 6)=0.82802.

Therefore, the minimum value of a is 6.

Hence, all the questions are answered.

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