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3 years ago

Jackson eats 1/4 part of a banana in 1/2 minute. How much time he will need to eat full banana?

1 answer:

4 0

Ignore the second fraction for a moment. say Jackson eats 1/4 of a banana every serving. how many servings would it take to eat the whole banana? 4, of course, because 4 1/4's equals one. now say that each serving is the same as 1/2 minute. Just multiply 1/2 minute by the 4 servings. 2 full minutes.

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At a local dealership 30% of cars are SUVs, 50% of the cars are painted white, and 15% are white and an SUV. If we randomly sele

alex41 [277]

Answer:

1/8, 0.0625 , or 6.25%

Step-by-step explanation:

50% of the cars are painted white.

1/2 * 1/2 = 1/4 * 1/4 = 1/8

Probability of A occuring 4 time(s) = 0.54 = 0.0625

Probability of A NOT occuring = (1 - 0.5)4 = 0.0625

Probability of A occuring = 1 - (1 - 0.5)4 = 0.9375

6 0

3 years ago

Find the two missing terms of the sequence and determine if the sequence is arithmetic, geometric, or neither.

wolverine [178]

-5, -7
The sequence is arithmetic.

6 0

3 years ago

On a number line, point D is at -3, and point E is at 6. The point F lies on DE. The ratio of DF to FE is 2:3. Where does point

expeople1 [14]

Answer: point F is at 0.6 on the number line

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Explanation:

The ratio 2:3 scales up to 2x:3x for some positive real number x.

This means the distance from D to F is 2x units, and the distance from F to E is 3x units. Combine those two smaller distances to get 2x+3x = 5x to represent the full distance from D to E.

D is at -3 and E is at 6. This is a distance of 9 units since |-3-6| = |-9| = 9

Set this equal to the 5x from earlier and solve

5x = 9

x = 9/5

x = 1.8

This leads to 2x = 2*1.8 = 3.6

Therefore, we'll move 3.6 units from -3 to -3+3.6 = 0.6 which is the location of point F on the number line.

Notice that from 0.6 to 6 is 5.4 units and that 3x = 3*1.8 = 5.4 matches up to help confirm the answer.

5 0

3 years ago

Use always, sometimes or never to make a true statement:

viva [34]

Answer:

1. Intersecting lines are always coplanar

2. Two planes never intersect in exactly one point

3. Three points are always coplanar

4. A plane containing two points of a line always contains the entire line

5. Four points are sometimes coplanar

6. Two lines never meet in more than one point.

7. Two skew lines are never coplanar

8. Line TQ and Line QT are always the same line.

Step-by-step explanation:

Note: Coplanar means "In the same plane"

1. Each line exist in many planes. But different lines must share at least one plane for them to intersect. That is why intersecting lines are always coplanar.

2. Two planes never intersect at exactly one point because only lines intersect at a point. Planes can only intersect along a line.

3.Three points are always coplanar because in geometry, a group of points are coplanar because there is a geometric plane that they all lie on.

4. A plane containing two points of a line always contains the entire line. Yes

5. Four points are only sometimes coplanar because there is a probability that they may all not lie on the same plane

6. Two lines never meet in more than one point because lines are basically straight and cannot bend over to intersect at another point

7. Two skew lines are never coplanar because skews lines are lines that do not intersect and are never parallel.

8. Line TQ and Line QT are always the same line because a line is straight and extends from one point to the other. So, if a line is labelled TQ calling it QT means the same thing

4 0

3 years ago

The n candidates for a job have been ranked 1, 2, 3,…, n. Let X 5 the rank of a randomly selected candidate, so that X has pmf p

Anon25 [30]

Answer:

A. E(x) = 1/n×n(n+1)/2

B. E(x²) = 1/n

Step-by-step explanation:

The n candidates for a job have been ranked 1,2,3....n. Let x be the rank of a randomly selected candidate. Therefore, the PMF of X is given as

P(x) = {1/n, x = 1,2...n}

Therefore,

Expectation of X

E(x) = summation {xP(×)}

= summation {X×1/n}

= 1/n summation{x}

= 1/n×n(n+1)/2

= n+1/2

Thus, E(x) = 1/n×n(n+1)/2

Value of E(x²)

E(x²) = summation {x²P(×)}

= summation{x²×1/n}

= 1/n

3 0

3 years ago