Help pleaseeeeeeeeeeeee

Oduvanchick [21]

This is a standard relationship/systems of equations question. Here is how you attack it. Firstly, set up equations to represent the relationships. Ed = 4*Kim That shows that Ed has four times as many pins as Kim. Next, we see that the two have 25 together. So: Ed + Kim = 25 Now we have our two equations. In order to solve, we need to get one of the equations down to one variable. We can achieve this via substitution. The first equation tells us that we can substitute 4*Kim for Ed (in the second equation). So, let's do just that: 4*Kim + Kim = 255*Kim = 25Kim = 5 Using the first equation again, we can solve for Ed: Ed = 4*KimEd = 4*(5)<span>Ed = 20</span>

6 0

3 years ago

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A student has a 30% chance of receiving chocolate ice cream for dessert at lunch. A statistics class designs a simulation to app

Nadusha1986 [10]

The count of the sections the class should label as "Chocolate Ice Cream" is 3.

<h3>How to calculate the probability of an event?</h3>

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

$P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}$

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

For the considered case, it is given that:

Probability of an student getting Chocolate Ice Cream for dessert at lunch = 30% = 0.3

The count of the sections the class should label as "Chocolate Ice Cream" should be such that spinner getting "Chocolate Ice Cream" as option should have the probability as 0.3

Let the count of the sections the class should label as "Chocolate Ice Cream" be $x$

Then, as there are in total 10 sections in the spinner, and all sections are assumingly equally probable, thus, if the event E is:

E = event of getting "Chocolate Ice Cream" in the spinner ,

then as n(E) = $x$ (as there are $x$ sections with label "Chocolate Ice Cream")

and n(S) = total count of sections (size of sample space) = 10

Thus, we get probability of event E as:

$P(E) = \dfrac{n(E)}{n(S)} = \dfrac{x}{10}$

This needs to be equal to 0.3, thus,

$0.3 = \dfrac{x}{10}\\x = 3$

Thus, the count of the sections the class should label as "Chocolate Ice Cream" is 3.

Learn more about probability here:

brainly.com/question/1210781

3 0

2 years ago

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BRAINLIEST if answered and up to 50 points. 1-3

Helen [10]

Answer:

are there any options or...

Step-by-step explanation:

4 0

3 years ago

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The pressure P exerted by a liquid on an object varies directly as the depth d of the object. If an object at depth 3m undergoes

GaryK [48]

3/485 = 7/ x

then 3x = 7*485 = 3395

then x = <span>1131.67 kg/m<span>2

Hope this helps! Have a great day!</span></span>

8 0

3 years ago

Help me pleaseeeeeeeeeeeeeee

boyakko [2]

Answer:

So the top right answer choice (x - 10)² = 80 has the same solution as x² - 20x + 20 = 0

Step-by-step explanation:

Let's solve your equation step-by-step.

x² − 20x + 20 = 0

For this equation: a = 1, b = -20, c = 20

1x² + −20x + 20 = 0

Step 1: Use quadratic formula with a = 1, b = -20, c = 20.

x = −b ± √b² − 4ac / 2a

x = −(−20) ± √(−20)² − 4(1)(20) / 2(1)

x = 20 ± √320 / 2

x = 10 + 4√5 or x = 10 − 4√5

Let's try the top right answer choice

(x−10)² = 80

Step 1: Simplify both sides of the equation.

x² − 20x + 100 = 80

Step 2: Subtract 80 from both sides.

x² − 20x + 100 − 80 = 80 − 80

x² − 20x + 20 = 0

For this equation: a = 1, b = -20, c = 20

1x² + −20x + 20 = 0

Step 3: Use quadratic formula with a = 1, b = -20, c = 20.

x = −b ± √b² − 4ac / 2a

x = −(−20) ± √(−20)² − 4(1)(20) / 2(1)

x = 20 ± √320 / 2

x = 10 + 4√5 or x = 10 − 4√5

6 0

2 years ago

Help pleaseeeeeeeeeeeee​

Help pleaseeeeeeeeeeeee