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2 years ago

12

1. Rich wants to buy a coat priced at x dollars. It is on sale for 20% off. He must

1 answer:

alexandr1967 [171]2 years ago

6 0

Answer: $x-0.20x+0.0725x$

Step-by-step explanation:

Let x be the price. 20% (0.20) 7.25% (0.0725)

$x-0.20x+0.0725x$

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Write an expression for the model. Find the sum.

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3 years ago

Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&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).

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3 years ago

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3 years ago

Mac earns 4927 bimonthly. how much does he earn per year

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3 years ago

Read 2 more answers

Trapezoids and kites

lisov135 [29]

QUESTION 3

The sum of the interior angles of a kite is $360\degree$ .

$\Rightarrow 36\degree +70\degree+m$ .

$\Rightarrow 106\degree+m$ .

$\Rightarrow m$ .

$\Rightarrow m$ .

But the two remaining opposite angles of the kite are congruent.

$\Rightarrow m$

$\Rightarrow m$ .

$\Rightarrow 2m$ .

$\Rightarrow m$ .

$\Rightarrow m$ .

QUESTION 4

RH is the hypotenuse of the right triangle formed by the triangle with side lengths, RH,12, and 20.

Using the Pythagoras Theorem, we obtain;

$|RH|^2=12^2+20^2$

$|RH|^2=144+400$

$|RH|^2=544$

$|RH|=\sqrt{544}$

$|RH|=4\sqrt{34}$

QUESTION 5

The given figure is an isosceles trapezium.

The base angles of an isosceles trapezium are equal.

Therefore $m$

QUESTION 6

The measure of angle Y and Z are supplementary angles.

The two angles form a pair of co-interior angles of the trapezium.

This implies that;

$m$

$\Rightarrow m$

$\Rightarrow m$

QUESTION 7

The sum of the interior angles of a kite is $360\degree$ .

$\Rightarrow 48\degree +110\degree+m$ .

$\Rightarrow 158\degree+m$ .

$\Rightarrow m$ .

$\Rightarrow m$ .

But the two remaining opposite angles are congruent.

$\Rightarrow m$

$\Rightarrow m$ .

$\Rightarrow 2m$ .

$\Rightarrow m$ .

$\Rightarrow m$ .

QUESTION 8

The diagonals of the kite meet at right angles.

The length of BC can also be found using Pythagoras Theorem;

$|BC|^2=4^2+7^2$

$\Rightarrow |BC|^2=16+49$

$\Rightarrow |BC|^2=65$

$\Rightarrow |BC|=\sqrt{65}$

QUESTION 9.

The sum of the interior angles of a trapezium is $360\degree$ .

$\Rightarrow m$ .

$\Rightarrow m$ .

But the measure of angle M and K are congruent.

$\Rightarrow m$ .

$\Rightarrow m$ .

$\Rightarrow m$ .

$\Rightarrow m$ .

6 0

3 years ago