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Alex Ar [27]
2 years ago
5

A mouse traveled a total distance of StartFraction 3 Over 24 EndFraction of a mile in a maze over the past 3 hours. The mouse tr

aveled the same distance each hour. To determine the distance that the mouse traveled each hour, Matt performed the calculations below. StartFraction 3 Over 24 EndFraction divided by 3 = StartFraction 3 Over (24 divided by 3) EndFraction = StartFraction 3 Over 8 EndFraction He concluded that the mouse traveled StartFraction 3 Over 8 EndFraction of a mile each hour. What is Matt's error?
Mathematics
2 answers:
Slav-nsk [51]2 years ago
8 0

Answer:

D.) Matt should have divided both 3 and 24 by 3.

Step-by-step explanation:

Artemon [7]2 years ago
3 0

Step-by-step explanation:

It is given that, a mouse covered a distance of 3/24 miles over a distance of 3 hours. Each hour, the mouse traveled the same distance. Distance traveled in 1 hour can be calculated by dividing 3/24 miles by 3. So,

d=\dfrac{\dfrac{3}{24}}{3}\\\\d=\dfrac{3}{24}\times \dfrac{1}{3}\\\\d=\dfrac{1}{24}\ \text{hours}

Matt calculations shows that the distance is 3/8 hours. While dividing 3/24 by 3, he doesn't take reciprocal of 3 as a result he get 9 on numerator and 24 on denominator. Simplified fraction of 9/24 is 3/8. But the correct answer should be 1/24 hours.

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In choice situations of this type, subjects often exhibit the "center stage effect," which is a tendency to choose the item in t
victus00 [196]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a

The probability that he or she would choose the pair of socks in the center position is   p =\frac{1}{5}

The correct answer choice is

X has a binomial distribution with parameters n=100 and p=1/5  

b

The mean is  \mu = 20

The standard deviation is \sigma=4

c

The probability, P =0.0002

d

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

Using the R the probability Pe = 0.0003

The probabilities P \approx Pe

Step-by-step explanation:

Since the person selects his or her desired pair of socks at random , then the probability that the person would choose the pair of socks in the center position from all the five identical pair is mathematically evaluated as

                  p =\frac{1}{5}

                    =0.2

The mean of this distribution is mathematical represented as

           \mu = np

substituting the value

         \mu = 100 * 0.2

             \mu = 20

The standard deviation is mathematically represented as

         \sigma = \sqrt{np (1-p)}

substituting the value

           = \sqrt{100 * 0,2 (1-0.2)}

           \sigma=4

Applying normal approximation the probability that 34 or more subjects would choose the item in the center if each subject were selecting his or her preferred pair of socks at random would be mathematically represented as

               P=P(X \ge 34 )

By standardizing the normal approximation we have that

              P(X \ge 34) \approx P(Z \ge z)

Now z is mathematically evaluated as

               z = \frac{x-\mu}{\sigma }

Substituting values

             z = \frac{34-20}{4}

               =3.5

So  using the z table the P(Z \ge 3.5) is  0.0002

The probability P and Pe that 34 or more subject would choose the center pair is very small  So

The correct answer is

The experiment supports the center stage effect. If participants were truly picking the socks at random, it would be highly unlikely for 34 or more to choose the center pair.

 

6 0
2 years ago
Round the number 76.4491 to 1 decimal place
solniwko [45]

Answer:

Step-by-step explanation:

76.4

4 0
2 years ago
An accident at an oil drilling platform is causing a circular-shaped oil slick to form. The volume of the oil slick is roughly g
Basile [38]

Answer:

V(t)=0.0112\pi t^{2}

Step-by-step explanation:

we have

V(r)=0.07\pi r^{2} -----> equation A

r(t)=0.4t -----> equation B

To find out (V of r)(t) substitute equation B in equation A

V(r(t))=V(t)

V(t)=0.07\pi (0.4t)^{2}

V(t)=0.07\pi (0.16)t^{2}

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What would you see at the wedding of Betty Crocker and the pillsbury dough boy
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A cement bridge post is 24 inches square and 15 feet 9 inches in length. If the cement weighs 145 pounds per cubic foot, how muc
lisov135 [29]

Answer:

The answer is option (C), One cement bridge post weighs 9,135 pounds

Step-by-step explanation:

Step 1: Get the total volume of a cement bridge post

The cement bridge post is in the shape of a cuboid, therefor the volume of the cement bridge post can be expressed as;

Volume of cement bridge post=Base area×Length

where;

Base area=(24^2) inches square

1 foot=12 inches

Convert 24 inches to foot=24/12=2^2=4 feet²

Length=15 feet and 9 inches

1 foot=12 inches

Convert 9 inches to foot=9/12=0.75 feet

Total length=(15+0.75)=15.75 feet

replacing;

Volume of cement bridge post=(4×15.75)=63 cubic feet

Volume of cement bridge post=63 cubic feet

Step 2: Get the total weight of the cement bridge post

Total weight of the cement bridge post=Weight per cubic foot×total volume of the cement bridge post

where;

Weight per cubic foot=145 pounds per cubic foot

Total volume of the cement bridge post=63 cubic feet

replacing;

Total weight of the cement bridge post=(145×63)=9,135 pounds

One cement bridge post weighs 9,135 pounds

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2 years ago
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