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Usimov [2.4K]
2 years ago
15

linda,rumi,liz,amy,adam,and sam ran a 800-meter race.Adam BEAT liz by 7meter Sam beat linda by 12 meters. Adam finished 5 meters

ahead of any but 3 meters behind Sam. Rumi finished halfway between the first and the last women. in what order did the women finish? what were the d I stances between them?
Mathematics
1 answer:
Stels [109]2 years ago
5 0
The distance between Adam and Liz is 7 meters. Sam ang Linda are 12 meters apart. Adam and Sam are 3 meters apart while Adam is 5 meters ahead of anyone. So, Sam finished first, then Adam, next is Liz, next is Rumi, last is Linda. 
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Which of the numbers below are some potential roots of p(x) = x3 + 6x2 − 7x − 60 according to the rational root theorem?
OlgaM077 [116]

Answer: -10,-5,3,15

A,C,D,E

The next one is a= -5, b=1, c=-5, d=-12

The last one is 3 & 4

4 0
2 years ago
There are 520 marbles in a jar. The ratio of red to yellow marbles is 3:4 and the ratio of yellow to blue marbles is 2:3. What i
frutty [35]

Answer:

Step-by-step explanation:

Change the 2/3 to 4/6

Now the ratio becomes 3:4:6

So the number of each is

3x + 4x + 6x = 520

13x = 520

x = 40

Red = 3*40 =      120

Yellow = 4*40 = 160

Blue = 6 * 40  = 240

Total =               520

7 0
2 years ago
find the probability that a randomly selected automobile tire has a tread life between 42000 and 46000 miles
maria [59]
Given that in a national highway Traffic Safety Administration (NHTSA) report, data provided to the NHTSA by Goodyear stated that the mean tread life of a properly inflated automobile tires is 45,000 miles. Suppose that the current distribution of tread life of properly inflated automobile tires is normally distributed with mean of 45,000 miles and a standard deviation of 2360 miles.

Part A:

Find the probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is between two numbers, a and b is given by:
P(a \ \textless \  X \ \textless \  b) = P(X \ \textless \  b) - P(X \ \textless \  a) \\  \\ P\left(z\ \textless \  \frac{b-\mu}{\sigma} \right)-P\left(z\ \textless \  \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life between 42,000 and 46,000 miles is given by:
P(42,000 \ \textless \ X \ \textless \ 46,000) = P(X \ \textless \ 46,000) - P(X \ \textless \ 42,000) \\ \\ P\left(z\ \textless \ \frac{46,000-45,000}{2,360} \right)-P\left(z\ \textless \ \frac{42,000-45,000}{2,360} \right) \\  \\ =P(0.4237)-P(-1.271)=0.66412-0.10183=\bold{0.5623}


b. Find the probability that randomly selected automobile tire has a tread life of more than 50,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, a, is given by:
P(X \ \textgreater \  a) = 1-P(X \ \textless \ a)  \\  \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of more than 50,000 miles is given by:
P(X \ \textgreater \  50,000) = 1 - P(X \ \textless \ 50,000) \\ \\ =1-P\left(z\ \textless \ \frac{50,000-45,000}{2,360} \right)=1-P(z\ \textless \ 2.1186) \\  \\ =1-0.98294=\bold{0.0171}


Part C:

Find the probability that randomly selected automobile tire has a tread life of less than 38,000 miles.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, a, is given by:
P(X \ \textless \  a) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles a standard deviation of 2360 miles.
The probability that randomly selected automobile tire has a tread life of less than 38,000 miles is given by:
P(X \ \textless \  38,000) = P\left(z\ \textless \ \frac{38,000-45,000}{2,360} \right) \\  \\ =P(z\ \textless \ -2.966)=\bold{0.0015}


d. Suppose that 6% of all automobile tires with the longest tread life have tread life of at least x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is greater than a numbers, x, is given by:
P(X \ \textgreater \ x) = 1-P(X \ \textless \ a) \\ \\ =1-P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at least x miles is 6%.

Thus:
P(X \ \textgreater \ x) =0.06 \\  \\ \Rightarrow1 - P(X \ \textless \ x)=0.06 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.06=0.94 \\  \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 1.555) \\ \\ \Rightarrow \frac{x-45,000}{2,360}=1.555 \\  \\ \Rightarrow x-45,000=2,360(1.555)=3,669.8 \\  \\ \Rightarrow x=3,669.8+45,000=48,669.8
Therefore, the value of x is 48,669.8


e. Suppose that 2% of all automobile tires with the shortest tread life have tread life of at most x miles. Find the value of x.
The probability that a normally distributed data set with a mean, μ, and standard deviation, σ, is less than a numbers, x, is given by:
P(X \ \textless \ x) =P\left(z\ \textless \ \frac{a-\mu}{\sigma} \right)
Given that the the mean tread life of a properly inflated automobile tires is 45,000 miles and a standard deviation of 2360 miles and that the probability that all automobile tires with the longest tread life have tread life of at most x miles is 2%.

Thus:
P(X \ \textless \ x)=0.02 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=1-0.02=0.98 \\ \\ \Rightarrow P\left(z\ \textless \ \frac{x-45,000}{2,360} \right)=P(z\ \textless \ 2.054) \\ \\ \Rightarrow \frac{x-45,000}{-2,360}=2.054 \\ \\ \Rightarrow x-45,000=-2,360(2.054)=-4,847.44 \\ \\ \Rightarrow x=-4,847.44+45,000=40,152.56
Therefore, the value of x is 40,152.56
4 0
2 years ago
Ten people can dig five holes in three hours. If n people digging at the same rate dig m holes in d hours:
Inessa [10]

First. get the unit rate. That is, transform "Ten people can dig five holes in three hours" into "One person digs one holes in one hour."

We can say that

\frac{5\,\mbox{holes}}{3\,\,\mbox{hours}\cdot\mbox{10\,\,\mbox{person}}}=\frac{\frac{5}{3}\,\,\mbox{holes}}{1\,\,\mbox{hour}\cdot\mbox{10\,\,\mbox{person}}}  

(that's still for 10 people). For 1 person this rate will be 1/10th of the above, so:

\frac{\frac{5}{3}\,\,\mbox{holes}}{1\,\,\mbox{hour}\cdot 10 \,\,\mbox{person}}=\frac{\frac{5}{30}\,\,\mbox{holes}}{1\,\,\mbox{hour}\cdot 1 \,\,\mbox{person}}

So the unit rate is 5/30. We can now set up an equation expressing the number of holes "m" dug up by "n" people in "d" hours:

m = \frac{5}{30}\cdot n\cdot d

which is clearly linear in n and d.

Now we can answer the questions:

(1) Is n proportional to m when d=3? Answer: Yes

m = \frac{5}{30}\cdot n \cdot 3 = \frac{1}{2}n\\\implies n = 2m

which shows that n is proportional to m in this case.

(2) Is n proportional to d when m=5? Answer: No

5 = \frac{5}{30}\cdot n \cdot d\\\implies n = \frac{30}{d}

There is an inverse proportionality, therefore this is not proportional.

(3) Is m proportional to d when n=10? Answer: Yes

m = \frac{5}{30}\cdot 10 \cdot d = \frac{5}{3}d

There is a proportional relationship between m and d.

3 0
2 years ago
The payout from a winning lottery varies inversely as the number of winners. If Alexis and her 7 co-workers each receive a $4.65
Masteriza [31]

Answer:

A) 12.4 million

Step-by-step explanation:

First you times the payment given out by the number of winners:

4.65 x 8 (don’t forget to count Alexis) = 37.2

Then, you’re left with the amount money won in total. You then divide that by the three people.

37.2/ 3 = 12.4

$12.4 million

3 0
2 years ago
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