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Ira Lisetskai [31]
2 years ago
9

A bag of candy contains 50 pieces of candy. The flavors are orange, strawberry, apple and

Mathematics
1 answer:
swat322 years ago
8 0

Answer:

15

Step-by-step explanation:

Let the probability of picking an orange = P(O), and the probability of picking a strawberry = P(S),

Based on the question, he picked an orange first and the probability of picking that is

P(O) = 12/50

Then he picked a strawberry on the second pick, the probability of picking the strawberry is P(S) and we'll find that later.

The probability of picking orange then strawberry with replacement is

P(O) × P(S) = 9/125

Substitute the P(O)=12/50,

12/50 × P(S) = 9/125

P(S) = 3/10

Then by finding the number of strawberry candies, we'll just have to multiply P(S) with the total number of candies, i.e.

Number of strawberry candy

= 3/10 × 50

= 15

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the answer is option B. angle S.

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A bike rental shop rents out bicycles with a flat rate of $6 and then charges $2 per hour a person rents out the bike. Identify
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Answer:

Independent Variable: Number of hours

Dependent Variable: Cost

Equation: C = 2x + 6

Step-by-step explanation:

    The independent variable (x) is the variable that is manipulated. In this instance, that would be the number of hours a person rents out the bike.

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A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the stor
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Answer:

Critical value is t = 1.9901

Step-by-step explanation:

We are given the results of the sampling :

                                     In-House Credit Card                National Credit Card                    <u>Sample Size</u> :                              32                                                    50

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<u>Standard Deviation </u>:                $10.90                                             $12.47

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<em>Firstly, we will specify our null and alternate hypothesis;</em>

Let \mu_1 = Mean Monthly Purchases of In-House Credit Card

     \mu_2 = Mean Monthly purchases of National Credit Card

So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0  {means that there is no difference in the mean monthly purchases by customers using the two types of credit cards}

Alternate Hypothesis, H_0 : \mu_1-\mu_2\neq 0  {means that there is statistical difference in the mean monthly purchases by customers using the two types of credit cards}

The test statistics that will be used here is <u>Two-sample t-test statistics</u>;

              T.S. = \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }  ~ t_n___1+n_2-_2

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So, degree of freedom of t-value here is (32 + 50 - 2) = 80

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<em>Therefore, the critical value assuming the population standard deviations are not known but that the populations are normally distributed with equal variances is t = 1.9901.</em>

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