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yuradex [85]
1 year ago
7

The price of a ring was increased by 8% to £486. What was the price before the increase?

Mathematics
1 answer:
viktelen [127]1 year ago
4 0

Answer:

38.88

Step-by-step explanation:

You might be interested in
Mario eats twice as many as Yolanda. If Yolanda eats 6 walnuts every day, how many walnuts does Mario eat in a week?
stepladder [879]

let

x-------> walnuts that Mario eats in a week

y-------> walnuts that Yolanda eats in a week


we know that

x=2y-------> equation 1

y=6*7-------> y=42 walnuts

substitute the value of y in equation 1

so

x=2*42-----> x=84 walnuts


therefore


the answer is

Mario eat in a week 84 walnuts

8 0
2 years ago
After being dropped from a platform, a ball bounces several times. The graph shows the height of the ball after each bounce.
Paul [167]

<u>Answer-</u>

<em>End behavior for increasing x represents that </em><em>the height of each bounce will approach 0.</em>

<u>Solution-</u>

From the graph the exponential equation is,

y=100e^{(-0.35x)}

From the properties of negative exponential function properties, as x increases, the value of y decreases.

So, in this case, as x or number of bounce increases, y or the height of bounce decreases. And eventually the value becomes zero.

Therefore, end behavior for increasing x represents that the height of each bounce will approach 0.

5 0
2 years ago
Orange M&amp;M’s: The M&amp;M’s web site says that 20% of milk chocolate M&amp;M’s are orange. Let’s assume this is true and set
SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let <em>X</em> = number of orange  milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, <em>p</em> = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, <em>n</em> = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable <em>X</em> follows a Binomial distribution with parameter <em>n</em> = 55 and <em>p</em> = 0.20.

The expected value of a Binomial random variable is:

E(X)=np

Compute the expected number of orange  milk chocolate M&M’s in a bag of 55 candies as follows:

E(X)=np

         =55\times 0.20\\=11

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}

                 =0.1968

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0
2 years ago
Write a quadratic function f whose zeros are -2 and 12.<br> f(x) = 0
borishaifa [10]

Step-by-step explanation:

f(x)=x²+10x-24

please rate and mark as brainliest

7 0
1 year ago
What is the product of StartFraction 4 Over 5 EndFraction times two-fifths? A model representing StartFraction 4 Over 5 EndFract
Anuta_ua [19.1K]

Answer:

StartFraction 8 Over 25

Step-by-step explanation:

What is the product of StartFraction 4 Over 5 EndFraction times two-fifths?

Mathematically =

StartFraction 4 Over 5 EndFraction = 4/5

Two- fifths = 2/5

Hence,

The product of StartFraction 4 Over 5 EndFraction times two-fifths

= 4/5 × 2/5

= 8/25

The correct option =

StartFraction 8 Over 25 EndFraction

3 0
2 years ago
Read 2 more answers
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