Match the events described to the likelihood of their occurrence.

Mathematics

1 answer:

Ksju [112]3 years ago

3 0

Answer:

Step-by-step explanation:

1. likelihood that edna picks a cherry.....impossible...because there is no cherries in the bag

2. communication....likelihood of an n is 2 / 13...15% chance...well...out of the answer choices, I am going with unlikely

3. mississippi....likelihood of s or p : 4/11 * 2/11 = 8/11...73% chance....its not certain...so I am going with neither likely or unlikely

4. all girls school....picking a girl is certain

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What is this area? please help

Gnoma [55]

Answer:

Step-by-step explanation:

multiply the 20m x 6m

20x6=120. Is base x high

then

divide ➗ by 2

is equal to 60

I hope help you

4 0

2 years ago

By showing your work what is the answer to 0.96÷0.144=

gayaneshka [121]

Recognize that both 0.96 and 0.144 are divisible by 12:

(0.96/12) / (0.144/12) = 0.08 / 0.012. This reduces to 0.02 / 0.003, or
20/3 or approx. 6.666.

You could also begin by eliminating the decimal fractions. Mult. 0.96 and 0.144 each by 1000 results in 960/144.

Since both 960 and 144 can be divided evenly by 24, we get 40 and 6.
40/6 = 20/3, or approx. 6.666, as before.

7 0

3 years ago

Almost all medical schools in the United States require students to take the Medical College Admission Test (MCAT). To estimate

yKpoI14uk [10]

Answer:

a) 0.5588

b) 0.9984

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 24

Standard Deviation, σ = 6.4

We are given that the distribution of score is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(score between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{6.4} \leq z \leq \displaystyle\frac{30-24}{6.4}) = P(-0.62 \leq z \leq 0.94)\\\\= P(z \leq 0.94) - P(z < -0.62)\\= 0.8264 - 0.2676 = 0.5588 = 55.88\%$

$P(20 \leq x \leq 30) = 55.88\%$

b) Sampling distribution

Sample size, n = 22

The sample will follow a normal distribution with mean 24 and standard deviation,

$s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{6.4}{\sqrt{22}} =1.36$

c) P(mean score of sample is between 20 and 30)

$P(20 \leq x \leq 30) = P(\displaystyle\frac{20 - 24}{1.36} \leq z \leq \displaystyle\frac{30-24}{1.36}) = P(-2.94 \leq z \leq 4.41)\\\\= P(z \leq 4.41) - P(z < -2.94)\\= 1 - 0.0016 = 0.9984 = 99.84\%$