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natulia [17]
2 years ago
7

A) Determine the functions whose graphs are parallel to the graph of the function y=0.5x+10 and intersect the graph of the funct

ion y= -1.5x. The functions are expressed by the formulas.
y = -1.5x + 6, y = 0.5x-6

b) Determine the functions whose graphs are parallel to the graph of the function y=0.5x+10 and intersect the graph of the function y= -1.5x. The functions are expressed by the formulas.

y = 0.5x+4, y = 0.5x, y = 3+1.5x
Mathematics
1 answer:
Elza [17]2 years ago
7 0

Answer:

y - 1 = 1/2(x -5)

Step-by-step explanation:

I don't know i am right tho so be careful

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Consider the inequality x + 7 < 9. Select all the true statements.​
alex41 [277]

Answer:

Step-by-step explanation:

here you go mate

step 1

x + 7 < 9  equation

step 2

x + 7 < 9  subtract -7

x+7(-7)<9(-7)

answer

x<2

3 0
3 years ago
One of the earliest applications of the Poisson distribution was in analyzing incoming calls to a telephone switchboard. Analyst
grandymaker [24]

Answer:

(a) P (X = 0) = 0.0498.

(b) P (X > 5) = 0.084.

(c) P (X = 3) = 0.09.

(d) P (X ≤ 1) = 0.5578

Step-by-step explanation:

Let <em>X</em> = number of telephone calls.

The average number of calls per minute is, <em>λ</em> = 3.0.

The random variable <em>X</em> follows a Poisson distribution with parameter <em>λ</em> = 3.0.

The probability mass function of a Poisson distribution is:

P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0,1,2,3...

(a)

Compute the probability of <em>X</em> = 0 as follows:

P(X=0)=\frac{e^{-3}3^{0}}{0!}=\frac{0.0498\times1}{1}=0.0498

Thus, the  probability that there will be no calls during a one-minute interval is 0.0498.

(b)

If the operator is unable to handle the calls in any given minute, then this implies that the operator receives more than 5 calls in a minute.

Compute the probability of <em>X</em> > 5  as follows:

P (X > 5) = 1 - P (X ≤ 5)

              =1-\sum\limits^{5}_{x=0} { \frac{e^{-3}3^{x}}{x!}} \,\\=1-(0.0498+0.1494+0.2240+0.2240+0.1680+0.1008)\\=1-0.9160\\=0.084

Thus, the probability that the operator will be unable to handle the calls in any one-minute period is 0.084.

(c)

The average number of calls in two minutes is, 2 × 3 = 6.

Compute the value of <em>X</em> = 3 as follows:

<em> </em>P(X=3)=\frac{e^{-6}6^{3}}{3!}=\frac{0.0025\times216}{6}=0.09<em />

Thus, the probability that exactly three calls will arrive in a two-minute interval is 0.09.

(d)

The average number of calls in 30 seconds is, 3 ÷ 2 = 1.5.

Compute the probability of <em>X</em> ≤ 1 as follows:

P (X ≤ 1 ) = P (X = 0) + P (X = 1)

             =\frac{e^{-1.5}1.5^{0}}{0!}+\frac{e^{-1.5}1.5^{1}}{1!}\\=0.2231+0.3347\\=0.5578

Thus, the probability that one or fewer calls will arrive in a 30-second interval is 0.5578.

5 0
3 years ago
Suppose that the relation G is defined as follows.
Serga [27]

G:{3,4,6}->{0,9}

The pairs represent the input (first number in each pair) and the result (second nu.ber in each pair) for the relation G. for example G(3)=9.

The domain is the set of values that the relation can act upon. The range is the set of the values the results can take

6 0
2 years ago
d) What are the values for coefficients a, b, and c in the quadratic equation for Daredevil Danny’s practice jump? (5 points)
SIZIF [17.4K]

Answer:

a = -0.32

b = 16

c = - 168

Step-by-step explanation:

In the figure attached, the quadratic function is shown. We can model it as follows:

y = a*(x - x1)*(x - x2)

where (x1, 0) and (x2, 0) are its roots

From the picture, roots are (15, 0) and (35, 0), replacing them into the equation:

y = a*(x - 15)*(x - 35)

and the vertex is at (25,32), replacing into the equation:

32 = a*(25 - 15)*(25 - 35)

32 = a*(-100)

a = 32/(-100)

a = -0.32

Applying distributive property:

y = -0.32*(x - 15)*(x - 35)

y = -0.32*(x² - 35x - 15x + 525)

y = -0.32x² + 16x - 168

which corresponds to the general form:

y = ax² + bx + c

3 0
3 years ago
Read 2 more answers
What is 3/5 divided by 7/9
Daniel [21]
3/5(divided)9/7
3/5x7/9
27/35
6 0
3 years ago
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