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erastova [34]
2 years ago
8

Plz help me plzzzzzzzzzz

Mathematics
1 answer:
Effectus [21]2 years ago
7 0

Answer:

I think the answer is 7;

Step-by-step explanation:

All i did was subtract Q11 from all 4 of its sides;

11=4=7

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The number of cases of a new disease can be modeled by the quadratic regression equation y = -2x^2 + 40x + 8. What is the best p
Stels [109]

Answer:

158 cases

Step-by-step explanation:

Given tbe quadratic regression model :

y = -2x^2 + 40x + 8

y = number of cases of a new disease

x = number of years

The predicted number of cases of a new disease in 15 years can be calculated thus ;

Put x = 15 in the equation ;

y = -2(15)^2 + 40(15) + 8

y = - 2 * 225 + 600 + 8

y = - 450 + 600 + 8

y = 158

158 cases

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2 years ago
Alex's wardrobe is 2 yards tall. How tall is the wardrobe in feet?<br> feet
hodyreva [135]
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How many terms are in the expression? 9a + 6b + 3c + 1 A) 1 B) 2 C) 3 D) 4
Kaylis [27]

Answer:

D.) there is 9a and 6b ,3c and 1 thats four different terms

Step-by-step explanation:

Have a nice day! :D

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2 years ago
Graph the system of equations. { x−y=6 4x+y=4 Use the Line tool to graph the lines.
lapo4ka [179]
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8 0
2 years ago
Suppose that, after measuring the duration of many telephone calls, a telephone company found their data was well-approximated b
Musya8 [376]

Answer:

a) 7.79%

b) 67.03%

c) Cumulative Distribution Function

P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b

Step-by-step explanation:

We are given the following in the question:

p(x) = 0.1 e^{-0.1x}

where x is the duration of a call, in minutes.

a) P( calls last between 2 and 3 minutes)

=\displaystyle\int^3_2 p(x)~ dx\\\\= \displaystyle\int^3_20.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^3_2\\\\=-\Big[e^{-0.3}-e^{-0.2}\Big]\\\\= 0.0779\\=7.79\%

b) P(calls last 4 minutes or more)

=\displaystyle\int^{\infty}_4 p(x)~ dx\\\\= \displaystyle\int^{\infty}_40.1e^{-0.1x}~dx\\\\=\Big[-e^{-0.1x}\Big]^{\infty}_4\\\\=-\Big[e^{\infty}-e^{-0.4}\Big]\\\\=-(0- 0.6703)\\= 0.6703\\=67.03\%

c) cumulative distribution function

P(t) = \displaystyle\int^{\infty}_{-\infty} 0.1e^{-0.1t}~dt\\\\= \displaystyle\int^{b}_{a} 0.1e^{-0.1t}~dt, ~~a\leq t \leq b

6 0
3 years ago
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