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Dennis_Churaev [7]
2 years ago
8

Four sisters bought a present for their mother. They received a 10% discount on the original price of the gift. After the discou

nt was taken, each sister payed $9. What was the original price of the gift
Mathematics
1 answer:
Orlov [11]2 years ago
4 0

Answer:

$40

Step-by-step explanation:

x is the original price of the gift

(10% of x) is the value of the discount

(10% of x) = 10/100 . x = 0.1 x

0.1 x    is the value of the discount

x - (10% of x) is what was paid, after the discount

x - 0.1 x = 0.9 x

0.9 x    is what was paid

4 children paid $9/each

4 times $9 = $36

$36 is the total amount that was paid by the children

$36 = 0.9 x

$36 / 0.9 = x

x = $36 / 0.9 = 360 / 9

x = 40

;-)

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At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t
grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c) The expected value of the temperature increase is 17.5°C.

Step-by-step explanation:

Let <em>X</em> = temperature increase.

The random variable <em>X</em> follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of <em>X</em> is:

f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

P(X

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

P(20

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable <em>X</em> as follows:

E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5

Thus, the expected value of the temperature increase is 17.5°C.

7 0
2 years ago
Aunt Andrea decided to help with decorations for Rachel’s party. Aunt Andrea created a table to find the amount of balloons she
mr Goodwill [35]

Answer:

y=5x

Step-by-step explanation:

Because if you look at it 5x any of them would equal the answer

5 0
2 years ago
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John has 36 sweets and he shares them in the ratio 2 : 7.
Lana71 [14]

Answer:

15 more

Step-by-step explanation:

John has 36 sweets and he shares them in the ratio 2 : 7.

How many sweets is the larger share?

7 0
2 years ago
Determine if each of the following sets is a subspace of ℙn, for an appropriate value of n. Type "yes" or "no" for each answer.
xxMikexx [17]

Answer:

1. Yes.

2. No.

3. Yes.

Step-by-step explanation:

Consider the following subsets of Pn given by

1.Let W1 be the set of all polynomials of the form p(t)=at^2, where a is in ℝ.

2.Let W2 be the set of all polynomials of the form p(t)=t^2+a, where a is in ℝ.

3. Let W3 be the set of all polynomials of the form p(t)=at^2+at, where a is in ℝ.

Recall that given a vector space V, a subset W of V is a subspace if the following criteria hold:

- The 0 vector of V is in W.

- Given v,w in W then v+w is in W.

- Given v in W and a a real number, then av is in W.

So, for us to check if the three subsets are a subset of Pn, we must check the three criteria.

- First property:

Note that for W2, for any value of a, the polynomial we get is not the zero polynomial. Hence the first criteria is not met. Then, W2 is not a subspace of Pn.

For W1 and W3, note that if a= 0, then we have p(t) =0, so the zero polynomial is in W1 and W3.

- Second property:

W1. Consider two elements in W1, say, consider a,b different non-zero real numbers and consider the polynomials

p_1 (t) = at^2, p_2(t)=bt^2.

We must check that p_1+p_2(t) is in W1.

Note that

p_1(t)+p_2(t) = at^2+bt^2  = (a+b)t^2

Since a+b is another real number, we have that p1(t)+p2(t) is in W1.

W3. Consider two elements in W3. Say p_1(t) = a(t^2+t), p_2(t)= b(t^2+t). Then

p_1(t) + p_2(t) = a(t^2+t) + b(t^2+t) = (a+b) (t^2+t)

So, again, p1(t)+p2(t) is in W3.

- Third property.

W1. Consider an element in W1 p(t) = at^2and a real scalar b. Then

bp(t) = b(at^2) = (ba)t^2).

Since (ba) is another real scalar, we have that bp(t) is in W1.

W3. Consider an element in W3 p(t) = a(t^2+t)and a real scalar b. Then

bp(t) = b(a(t^2+t)) = (ba)(t^2+t).

Since (ba) is another real scalar, we have that bp(t) is in W3.

After all,

W1 and W3 are subspaces of Pn for n= 2

and W2 is not a subspace of Pn.  

6 0
2 years ago
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kondaur [170]
\bf (x)(x+2)=255\implies x^2+2x=255\implies x^2+2x-255=y&#10;\\\\&#10;\textit{setting y=0}\implies x^2+2x-255=0&#10;\\\\&#10;\textit{now, factoring that}&#10;\\\\&#10;(x+17)(x-15)=0\implies &#10;\begin{cases}&#10;x=-17\\&#10;x=15&#10;\end{cases}

notice the picture of the graph added here
low and behold, x = -17, y is 0, and x= 15, y is 0
the graph is touching the x-axis, an x-intercept
or so-called, a "solution"

3 0
1 year ago
Read 2 more answers
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