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NNADVOKAT [17]
2 years ago
8

The course length of the Boston Marathon is about 4.2×104 meters. The course length of the Iditarod Trail Invitational ultra-mar

athon is about 1.6×106 meters. About how many times as long is the course of the Iditarod Trail ultra-marathon as that of the Boston Marathon?
Mathematics
1 answer:
GREYUIT [131]2 years ago
5 0

Answer:

The ultra-marathon Idita Rod Trail is about 38 times longer than the Boston Marathon.

Step-by-step explanation:

Let

x -----> the length of the Boston Marathon in meters

y -----> the  length of the Idita rod Trail Invitational ultra-marathon in meters

we have

x=4.2*10^4\ m

y=1.6*10^6\ m

we know that

To find out how many times as long is the course of the Idita rod Trail ultra-marathon as that of the Boston Marathon, divide the length of the Idita rod Trail Invitational ultra-marathon by the length of the  Boston Marathon

so

\frac{y}{x}

\frac{x}{y}=\frac{1.6*10^6}{4.2*10^4}

Remember that

To divide two numbers in scientific notation, divide their coefficients and subtract their exponents

\frac{1.6*10^6}{4.2*10^4}=\frac{1.6}{4.2}*(10^{6-4})=0.381*(10^{2})=38.1

therefore

The ultra-marathon Idita Rod Trail is about 38 times longer than the Boston Marathon.

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Harrison Water Sports has two retail outlets: Seattle and Portland. The Seattle store does 60 percent of the total sales in a ye
Anna11 [10]

Answer: P = 0.75

Step-by-step explanation:

Hi!

The sample space of this problems is the set of all the possible sales. It is divided in the disjoint sets:

S_s = {\text{sales made in Seattle }}\\S_p={\text {sales made in Portland}}

We have also the set of sales of boat accesories S_b, the colored one in the image.

We are given the data:

P(S_s) = 0.6\\P(S_b | S_s) = \frac{P(S_b\bigcap S_s)}{P(S_s)}=0.4\\P(S_b|S_p) =\frac{P(S_b\bigcap S_p)}{P(S_p)}=0.2

From these relations you can compute the probabilities of the intersections colored in the image:

pink\;set:\;P(S_b \bigcap S_s) =0.6*0.4=0.24\\blue\;set\;:P(S_b \bigcap S_p)=(1-0.6)*0.2 =0.08

You are asked about the conditional probability:

P(S_s|S_b) = \frac{P(S_s \bigcap S_b)}{P(S_b)}

To calculate this, you need  P(S_b) . In the image you can see that the set S_b is the union of the two disjoint pink and blue sets. Then:

P(S_b)=P((S_b \bigcap S_s)\bigcup(S_b \bigcap S_p)) = 0.24 + 0.08 = 0.32

Finally:

P(S_s|S_b) = \frac{0.24}{0.32}=0.75

4 0
2 years ago
Which statement is true about the relationship between the amount of plant food remaining and the number of days? Can you help m
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A contractor has submitted bids on three state jobs: an office building, a theater, and a parking garage. State rules do not all
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Answer:

The following are the answer to this question:

Step-by-step explanation:

In the given question the numeric value is missing which is defined in the attached file please fine it.  

Calculating the probability of the distribution for x:

\to f(x) = 0.19\  for \ x=14\\\\\to  f(x) = 0.29 \ for\ x=7\\\\\to f(x) = 0.38\  for \ x=1\\\\\to f(x)=0.14 \ for \ x=0\\

The formula for calculating the mean value:

\bold{ E(X)= x \times f(x)}

          =14 \times 0.19+7 \times 0.29+1 \times 0.38+0\times 0.14\\\\=2.66 + 2.03+0.38+ 0\\\\=5.07

\bold{E(X^2) = x^2 \times f(x)}

           =14^2 \times 0.19+7^2 \times 0.29+1^2 \times 0.38+0^2 \times 0.14 \\\\=196 \times 0.19+ 49 \times 0.29+1 \times 0.38+0 \times 0.14\\\\= 37.24+ 14.21+ 0.38+0 \\\\=51.83

use formula for calculating the Variance:

\to \bold{\text{Variance}= E(X^2) -[E(X)]^2}

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calculating the value of standard deivation:

Standard Deivation (SD) = \sqrt{Variance}

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6 0
2 years ago
Suppose that the weight of seedless watermelons is normally distributed with mean 6.1 kg. and standard deviation 1.9 kg. Let X b
horrorfan [7]

Answer:

a. X\sim N(\mu = 6.1, \sigma = 1.9) b. 6.1 c. 0.6842 d. 0.4166 e. 0.1194 f. 8.5349

Step-by-step explanation:

a. The distribution of X is normal with mean 6.1 kg. and standard deviation 1.9 kg. this because X is the weight of a randomly selected seedless watermelon and we know that the set of weights of seedless watermelons is normally distributed.

b. Because for the normal distribution the mean and the median are the same, we have that the median seedless watermelong weight is 6.1 kg.

c. The z-score for a seedless watermelon weighing 7.4 kg is (7.4-6.1)/1.9 = 0.6842

d. The z-score for 6.5 kg is (6.5-6.1)/1.9 = 0.2105, and the probability we are seeking is P(Z > 0.2105) = 0.4166

e. The z-score related to 6.4 kg is z_{1} = (6.4-6.1)/1.9 = 0.1579 and the z-score related to 7 kg is z_{2} = (7-6.1)/1.9 = 0.4737, we are seeking P(0.1579 < Z < 0.4737) = P(Z < 0.4737) - P(Z < 0.1579) = 0.6821 - 0.5627 = 0.1194

f. The 90th percentile for the standard normal distribution is 1.2815, therefore, the 90th percentile for the given distribution is 6.1 + (1.2815)(1.9) = 8.5349

7 0
2 years ago
A calculator has 50 keys in five colors:gray,black,blue,yellow,and green.There are 6 gray keys for every 7 blue keys.write the p
Akimi4 [234]
A ratio is written in the form a : b. In this case, there are 6 gray keys for every 7 blue keys. So the ratio for gray to blue keys would be 6:7. 
Since 6 and 7 don't have common factors, the ratio 6:7 cannot be simplified further. 
So the ratio for gray to blue keys is 6:7.

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2 years ago
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