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2 months ago

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A jar of sweets contains 5 yellow sweets, 4 red sweets, 8 green sweets, 4 orange sweets and 3 white sweets. Ola chooses a sweet

at random. What is the probability that she will pick either a yellow or orange sweet?

1 answer:

Zina [12.3K]2 months ago

7 0

Answer:

The probability of selecting either yellow or orange sweets is 3 / 8

Step-by-step explanation:

Given:

Count of yellow sweets = 5

Count of red sweets = 4

Count of green sweets = 8

Count of orange sweets = 4

Count of white sweets = 3

Find:

The chance of choosing a yellow or orange sweet

Computation:

Total number of sweets = 24

The probability for choosing a yellow sweet is 5 / 24

The probability for selecting an orange sweet is 4 / 24

The probability of selecting either yellow or orange sweets can be calculated as 5/24 + 4 / 24

This gives us a probability of 9 / 24 for either yellow or orange sweets.

The probability of selecting either yellow or orange sweets is 3 / 8

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A child rides a pony on a circular track whose radius is 5.25 m. a) Find the distance traveled and the displacement after the ch

tester [12383]

Answer:

a) The distance covered halfway = 16.49 [m], while the displacement at that point = 10.5 [m]b)

The distance traveled increases

when the child finishes one lap around the track.

Step-by-step explanation:

a)

Distance is a scalar measurement

that indicates how much ground an object has traversed in motion. Contrarily, displacement is a vector measurement which describes how far an object is from its original position after moving.

In this scenario, the child travels halfway around the circular track from point 1 to point 2. We know that the circumference or perimeter of a circle can be calculated as 2*PI*radius. Since the child only covers half the distance, we need to divide the circumference by 2, leading to distance = PI*radius

The halfway distance = (PI)*(5.25) = 16.49 [m]

Considering the definition of displacement, it’s clear that the distance from point 1 to point 2 equals one diameter from the starting point:

The halfway displacement = D = 2*radius = 2(5.25) = 10.5 [m]

b) What occurs to the distance after the child completes one full circuit? The distance will increase since the child covers more ground during the full lap:

The distance for one circuit = 2*(PI)*radius = 2(PI)(5.25) = 32.99 [m]

32.99 is greater than 16.49, confirming that the distance does indeed increase.

3 0

1 month ago

A quadratic function has a line of symmetry at x = –3.5 and a zero at –9. What is the distance from the given zero to the line o

Svet_ta [12734]

Given:

A quadratic function has a line of symmetry positioned at x = –3.5 with one root located at –9.

To find:

The second root.

Solution:

It is understood that the line of symmetry splits the quadratic function's graph into two identical halves. Hence, both roots are equidistant from this line.

This implies that the line of symmetry passes through the midpoint of the two roots.

Let the other root be denoted as x.

$-3.5=\dfrac{(-9)+x}{2}$

Multiply both sides by 2.

$-7=-9+x$

Add 9 to both sides.

$-7+9=-9+x+9$

$2=x$

Consequently, the other zero of the quadratic function is concluded to be 2.

8 0

1 month ago

What are the solutions of the equation x4 + 6x2 + 5 = 0? Use u substitution to solve.

tester [12383]

The solutions to the equation are $x=\pm i, x=\pm \sqrt{5}i$

Explanation:

Given the equation $x^{4}+6 x^{2}+5=0$

, we aim to find the solutions of the equation.

Let us use $x^{2} =u$ and $x^4=u^2$

Thus, the equation transforms into

$u^{2}+6 u+5=0$

Factoring the equation results in:

$(u+1)(u+5)=0$

$u=-1, u=-5$

Inserting back $x^{2} =u$ and allows us to solve for x.

Initially, we will substitute $u=-1$

Consequently, we find;

$x^{2} =-1$

$x=\sqrt{-1}$

$x=\pm i$

Similarly, by substituting $u=-5$ , we derive;

$x^{2} =-5$

$x=\sqrt{-5}$

$x=\pm \sqrt{5}i$

Thus, the solutions to the equation are $x=\pm i, x=\pm \sqrt{5}i$

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22 days ago

Mrs.williams is organizing her office supplies. There are 3 open boxes of paper clips in her desk drawer.each box has 7/8 of pap

Zina [12379]

21/8 or 2 and 5/8 boxes remaining

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20 days ago

Suppose we are given 4 sets A, B, C, D such that A ⊆ B and C ⊆ D such that A and C have no elements in common. Prove or give a c

babunello [11817]

Here’s a counterexample: consider

$B = \{1, 2, 3, 4, 5\},\quad D = \{A, B, C, D, 5\}$

Select the subsets in the following manner:

$A = \{1, 5\},\quad C = \{A, B, C\}$

It's accurate that $A\subseteq B$ and $C\subseteq D$ and that $A\cap C=\emptyset$ , but $A\cap D = \{5\}$

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1 month ago