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3 years ago

a store has 300 televisions on order, and 80% are high definition. how many televisions on order are HD?

Mathematics

2 answers:

Korolek [52]3 years ago

8 0

240 Televisions are on HD.

Dafna1 [17]3 years ago

5 0

Answer: The number of televisions on order are HD = 240

Step-by-step explanation:

Given : The number of televisions on order= 300

The percentage of televisions orders are high definition = 80%

i..e The proportion of televisions orders are high definition = 0.80 [ we divide percentage by 100 to convert into decimal]

Now , the number of televisions on order are HD = (Total televisions) x (proportion of televisions are high definition)

=(300) x (0.80)

=240

Hence, the number of televisions on order are HD = 240

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3 years ago

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3 years ago

There is 1/4 ounce of yeast in every 2 1/4 teaspoon of yeast. A recipe calls for 2 teaspoons of yeast. How many ounces of yeast

MrRissso [65]

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3 years ago

Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi

ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles carry just one person.

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

$P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}$

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

$P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291$

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

$\mu = np = 92(0.64) = 58.88$

$\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60$

We have to evaluate the probability that more than 47 cars carry just one person.

$P(x \geq 47)$

After continuity correction, we will evaluate

$P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)$

$= 1 - P(z \leq -2.6913)$

Calculation the value from standard normal z table, we have,

$P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%$

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0

3 years ago