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2 years ago

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:

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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Answer:

the answer is option B. angle S.

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3 years ago

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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.

The dependent variable (C) is the variable that is manipulated depending on the value of the independent variable. In this instance, that would be the total cost for renting the bike.

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3 years ago

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 Sample Size : 32 50

Mean Monthly Purchases : $45.67 $39.87

Standard Deviation : $10.90 $12.47

Also, the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level α of 0.05.

Firstly, we will specify our null and alternate hypothesis;

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 Two-sample t-test statistics;

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

where, $\bar X_1$ = Sample mean Purchases of In-House Credit Card = $45.67

$\bar X_2$ = Sample mean Purchases of National Credit Card = $39.87

$s_p$ = pooled variance

$n_1$ = In-house credit card sample = 32

$n_2$ = National credit card sample = 50

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

Now, at 0.05 significance level, t table gives critical value of t = 1.9901 at 80 degree of freedom.

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.

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2 years ago

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Answer:

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