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2 years ago

5

Problem

1 answer:

gregori [183]2 years ago

6 0

Answer:

62.5%

Step-by-step explanation:

150/240

Divide top and bottom by 10

15/24

Divide by 3

5/8

.625

Multiply by 100%

62.5%

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How can credit cards be safer than cash? a. It is harder to lose a single credit card than a large number of bills. b. A credit

konstantin123 [22]

Answer: Due to its anonymous nature, if you lose your cash, it's gone. Payment cards can be cancelled and replaced swiftly and remotely. Whilst cash may offer you complete anonymity from data thieves, EMV chip enabled cards offer a more thorough form of payment authentication, providing an extra layer of payment security

So, therefore, C.

3 0

3 years ago

A fair coin is flipped twelve times. What is the probability of the coin landing tails up exactly nine times?

seraphim [82]

Answer:

$P\left(E\right)=\frac{55}{1024}$

Step-by-step explanation:

Given that a fair coin is flipped twelve times.

It means the number of possible sequences of heads and tails would be:

2¹² = 4096

We can determine the number of ways that such a sequence could contain exactly 9 tails is the number of ways of choosing 9 out of 12, using the formula

$nCr=\frac{n!}{r!\left(n-r\right)!}$

Plug in n = 12 and r = 9

$=\frac{12!}{9!\left(12-9\right)!}$

$=\frac{12!}{9!\cdot \:3!}$

$=\frac{12\cdot \:11\cdot \:10}{3!}$ ∵ $\frac{12!}{9!}=12\cdot \:11\cdot \:10$

$=\frac{1320}{6}$ ∵ $3!\:=\:3\times 2\times 1=6$

$=220$

Thus, the probability will be:

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$

$=\frac{220}{4096}$

$=\frac{55}{1024}$

Thus, the probability of the coin landing tails up exactly nine times will be:

$P\left(E\right)=\frac{55}{1024}$

4 0

3 years ago

Part 2 of Question please help me <br> and thank you for your help

san4es73 [151]

1 tick represents 8 units

4 0

1 year ago

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist

Tcecarenko [31]

Answer:

a. 0.7291

b. 0.9968

c. 0.7259

Step-by-step explanation:

a. np and n(1-p) can be calculated as:

$np=23\times 0.48\\\\=11.04\\\\n(1-p)=23(1-0.52)\\\\=11.96$

#Both np and np(1-p) are greater than 5, hence, normal approximation is most appropriate:

$\mu_x=11.04\\\\\sigma^2=np(1-p)=0.48\times 0.52\times 23=5.7408$

#Define Y:

Y~(11.04,5.7408)

$P(X\leq 12)\approx P(Y\leq 12.5)\\\\P(Z\leq \frac{(12.5-11.04)}{\sqrt{5.7408}})=\\\\=1-0.2709\\\\=0.7291$

Hence, the probability of 12 or fewer is 0.8291

b. The probability that 5 or more fish were caught.

#Using normal approximation:

$P(X \geq 5) \approx P(Y \geq 4.5) = P(Z \geq\frac{ (4.5-11.04)}{\sqrt{(5.7408)}})\\\\=1-0.0032\\\\=0.9968$

Hence, the probability of catching 5+ is 0.9968

c. The probability of between 5 and 12 is calculated as;

-From b above $P(X\geq 5)=0.9968$ and a , $P(X\leq 12)$ =0.7291

$P(5\leq X\leq 120\approx P(4.5\leq Y\leq 12.5)\\\\=0.7291-(1-0.9968)\\\\=0.7259$

Hence, the probability of between 5 and 12 is 0.7259

4 0

3 years ago

Channing jogs 10 miles a week. How many miles will she jogot in 52 weeks?

Alborosie

It might me this.. 10 * 52 = 520

6 0

3 years ago