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MrRissso [65]
2 years ago
14

A large bag of marbles contains an equal amount of red, orange, yellow, blue, and green marbles. Quinn randomly selects 1 marble

from the bag. Which of the following represents the probability of selecting a red or blue marble?
Mathematics
1 answer:
Evgesh-ka [11]2 years ago
8 0
Since there are equal of each and there are four colors it is out of four, now because it is an equal amount of each color and there are two colors then there is a 2/4 chance of pulling a red or blue marble.
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Wilson has a balance of $890 on a credit card with an APR of 18.7%, compounded monthly. About how much will he save in interest
Leona [35]
890×(1+0.187÷12)^(12)−890×(1+0.125÷12)^(12)
=63.61 saved
8 0
1 year ago
Read 2 more answers
Two bicyclists are 7/8 of the way through a mile long tunnel when a train approaches the closer end at 40 mph. The riders take o
Anastaziya [24]

Answer:

the cyclists rode at 35 mph

Step-by-step explanation:

Assuming that the cyclists stopped, and accelerated instantaneously at the same speed than before but in opposite direction , then

distance= speed*time

since the cyclists and the train reaches the end of the tunnel at the same time and denoting L as the length of the tunnel :

time = distance covered by cyclists / speed of cyclists = distance covered by train / speed of the train

thus denoting v as the speed of the cyclists :

7/8*L / v = L / 40 mph

v = 7/8 * 40 mph = 35 mph

v= 35 mph

thus the cyclists rode at 35 mph

6 0
2 years ago
If the code for CAB is DEK, what is the code for BED?
snow_tiger [21]

Answer:

CIM

Step-by-step explanation:

C is the 3rd letter of the alphabet, A is the 1st, and B is the 2nd.

CAB = 3,1,2

Repeating for DEK:

DEK = 4,5,11

Comparing:

4−3 = 1

5−1 = 4

11−2 = 9

BED = 2,5,4, so adding the corresponding numbers:

2+1 = 3

5+4 = 9

4+9 = 13

So the code is CIM.

7 0
2 years ago
The Graduate Management Admission Test (GMAT) is a standardized exam used by many universities as part of the assessment for adm
Nat2105 [25]

Answer:

a) 16% of GMAT scores are 647 or higher.

b) 2.5% of GMAT scores are 647 or higher.

c) 34% of GMAT scores are between 447 and 547.

d) 81.5% of GMAT scores are between 347 and 647.

Step-by-step explanation:

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

Mean = 547

Standard deviation = 100

a. What percentage of GMAT scores are 647 or higher?

The Empirical rule states that 68% of the scores are within 1 standard deviation of the mean, that is, from 547 - 100 = 447 to 547 + 100 = 647. So 32% of the scores are outside the interval. Since the distribution is symmetric, 16% of them are lower than 447 and 16% of them are higher than 647.

So

16% of GMAT scores are 647 or higher.

b. What percentage of GMAT scores are 747 or higher (to 1 decimal)?

The Empirical rule states that 95% of the scores are within 2 standard deviations of the mean, that is, from 547 - 2*347 = 347 to 547 + 2*100 = 747. So 5% of the scores are outside the interval. Since the distribution is symmetric, 2.5% of them are lower than 347 and 2.5% of them are higher than 757

So

2.5% of GMAT scores are 647 or higher.

c. What percentage of GMAT scores are between 447 and 547?

447 is one standard deviation below the mean. The Empirical rule states that 68% of the scores are within 1 standard deviation of the mean, and since the distribution is symmetric, 34% are within one standard deviation below the mean and the mean, and 34% are within the mean and one standard deviation above the mean.

547 is the mean

447 is one standard deviation below the mean

So 34% of GMAT scores are between 447 and 547.

d. What percentage of GMAT scores are between 347 and 647 (to 1 decimal)?

The easist way is adding the percentage of scores from 347 to the mean(547) and the mean to 647.

Between 347 and 547

347 is two standard deviations below the mean. The Empirical rule states that 95% of the scores are within 2 standard deviations of the mean, and since the distribution is symmetric, 47.5% are within two standard deviation below the mean and the mean, and 47.5% are within the mean and two standard deviations above the mean.

So 47.5% of the scores are between 347 and 547

Between 547 and 647

447 is one standard deviation above the mean. The Empirical rule states that 68% of the scores are within 1 standard deviation of the mean, and since the distribution is symmetric, 34% are within one standard deviation below the mean and the mean, and 34% are within the mean and one standard deviation above the mean.

So 34% of the scores are between 547 and 647.

Between 347 and 647

47.5 + 34 = 81.5% of GMAT scores are between 347 and 647.

7 0
2 years ago
A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $1.10; if
topjm [15]

Answer:

a) The expected value is \frac{-1}{15}

b) The variance is  \frac{49}{45}

Step-by-step explanation:

We can assume that both marbles are withdrawn at the same time. We will define the probability as follows

#events of interest/total number of events.

We have 10 marbles in total. The number of different ways in which we can withdrawn 2 marbles out of 10 is \binom{10}{2}.

Consider the case in which we choose two of the same color. That is, out of 5, we pick 2. The different ways of choosing 2 out of 5 is \binom{5}{2}. Since we have 2 colors, we can either choose 2 of them blue or 2 of the red, so the total number of ways of choosing is just the double.

Consider the case in which we choose one of each color. Then, out of 5 we pick 1. So, the total number of ways in which we pick 1 of each color is \binom{5}{1}\cdot \binom{5}{1}. So, we define the following probabilities.

Probability of winning: \frac{2\binom{5}{2}}{\binom{10}{2}}= \frac{4}{9}

Probability of losing \frac{(\binom{5}{1})^2}{\binom{10}{2}}\frac{5}{9}

Let X be the expected value of the amount you can win. Then,

E(X) = 1.10*probability of winning - 1 probability of losing =1.10\cdot  \frac{4}{9}-\frac{5}{9}=\frac{-1}{15}

Consider the expected value of the square of the amount you can win, Then

E(X^2) = (1.10^2)*probability of winning + probability of losing =1.10^2\cdot  \frac{4}{9}+\frac{5}{9}=\frac{82}{75}

We will use the following formula

Var(X) = E(X^2)-E(X)^2

Thus

Var(X) = \frac{82}{75}-(\frac{-1}{15})^2 = \frac{49}{45}

7 0
2 years ago
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