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frutty [35]
3 years ago
7

"Mia jogs 3 kilometers in 20 minutes. There are about 0.6 miles in a kilometer. What is Mia’s approximate speed in miles per min

ute?
A. 0.25 miles per minute
B. 0.09 miles per minute
C. 4 miles per minute
D. 11 miles per minute"
Mathematics
2 answers:
Svetach [21]3 years ago
8 0

Answer:

B

Step-by-step explanation:

I smartest I'm know its B

Lapatulllka [165]3 years ago
4 0
First we need to find Mia's speed (v) in km/min

v = distance travelled/ time 
v = 3 km/20 minutes
v = 0.15 km /min

in order to convert the speed into mile/min, we need to use the conversion factor given in which for every kilometer, there is 0.6 miles. 

v = 0.15 km/min *(0.6 miles/ km) 
v =0.09 miles/min

therefore the speed in miles/min is 0.09 miles/min
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Answer:

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3 0
3 years ago
Determine if the sequence below is arithmetic or geometric and determine the
asambeis [7]

Answer:

Step-by-step explanation:

This is an arithmetic series.

3 , 5 , 7 ,......

Common difference = second term - first term

                                 = 5 - 3

                                  = 2

4 0
3 years ago
A quiz-show contestant is presented with two questions, question 1 and question 2, and she can choose which question to answer f
Mrrafil [7]

Answer:

The contestant should try and answer question 2 first to maximize the expected reward.

Step-by-step explanation:

Let the probability of getting question 1 right = P(A) = 0.60

Probability of not getting question 1 = P(A') = 1 - P(A) = 1 - 0.60 = 0.40

Let the probability of getting question 2 right be = P(B) = 0.80

Probability of not getting question 2 = P(B') = 1 - P(B) = 1 - 0.80 = 0.20

To obtain the better option using the expected value method.

E(X) = Σ xᵢpᵢ

where pᵢ = each probability.

xᵢ = cash reward for each probability.

There are two ways to go about this.

Approach 1

If the contestant attempts question 1 first.

The possible probabilities include

1) The contestant misses the question 1 and cannot answer question 2 = P(A') = 0.40; cash reward associated = $0

2) The contestant gets the question 1 and misses question 2 = P(A n B') = P(A) × P(B') = 0.6 × 0.2 = 0.12; cash reward associated with this probability = $200

3) The contestant gets the question 1 and gets the question 2 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.4×0) + (0.12×200) + (0.48×300) = $168

Approach 2

If the contestant attempts question 2 first.

The possible probabilities include

1) The contestant misses the question 2 and cannot answer question 1 = P(B') = 0.20; cash reward associated = $0

2) The contestant gets the question 2 and misses question 1 = P(A' n B) = P(A') × P(B) = 0.4 × 0.8 = 0.32; cash reward associated with this probability = $100

3) The contestant gets the question 2 and gets the question 1 too = P(A n B) = P(A) × P(B) = 0.6 × 0.8 = 0.48; cash reward associated with this probability = $300

Expected reward for this approach

E(X) = (0.2×0) + (0.32×100) + (0.48×300) = $176

Approach 2 is the better approach to follow as it has a higher expected reward.

The contestant should try and answer question 2 first to maximize the expected reward.

Hope this helps!!!

3 0
3 years ago
Sallie Woo worked these hours last week: Monday, 10; Tuesday, 8 1/4; Wednesday, 9 1/2; Thursday, 6; Friday, 8; Saturday, 5. Sall
almond37 [142]

Answer:

Her pay for the week= $ 611.325

Step-by-step explanation:

Sallie is paid $ 11.40 per hour for regular hours.

The total regular hours worked during the week are

Monday + Tuesday + Wednesday + Thursday+ friday = 8 + 8+ 8+6+8= 38 hours

She works overtime during the week

Monday + Tuesday + Wednesday = 2+ 1/4+  1 1/2=  2+ 1/4 + 3/2

= 8 +1+ 6/4=15/4 = 3 3/4 hours

Over the weekend she works five hours .

So the total pay would be

$ 11.40( 38 ) + $ 11.40( 1.5) (  15/4) + $ 11.40 ( 2) ( 5)

= 433.2 + 64.125 + 114.0

= $ 611.325

3 0
3 years ago
Haley works at a candy store. There are 10 types of bulk candy. Find the probability that one type of candy will be chosen more
DIA [1.3K]

Answer:

Probability that one type of candy will be chosen more than once in 10 trials = 0.2639

Step-by-step explanation:

This is a binomial experiment because

- A binomial experiment is one in which the probability of success doesn't change with every run or number of trials.

- It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure. (10 trials, with the outcome of each trial being that we get the required candy or not)

- The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 10 trials

x = Number of successes required = number of times we want to pick a particular brand of candy = more than once, that is > 1

p = probability of success = probability of picking a particular brand of candy from a bulk with 10 different types of candies = (1/10) = 0.10

q = probability of failure = Probability of not picking our wanted candy = 1 - p = 1 - 0.1 = 0.90

P(X > 1) = 1 - P(X ≤ 1)

P(X ≤ 1) = P(X = 0) + P(X = 1)

P(X = 0) = ¹⁰C₀ (0.10)⁰ (0.90)¹⁰⁻⁰ = 0.3486784401

P(X = 1) = ¹⁰C₁ (0.10)¹ (0.90)¹⁰⁻¹ = 0.387420489

P(X ≤ 1) = 0.3486784401 + 0.387420489 = 0.7360989291

P(X > 1) = 1 - 0.7360989291 = 0.2639010709 = 0.2639

Hope this Helps!!!

5 0
3 years ago
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