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3 years ago

Factor the quadratic expression in the equation y = 2x2 + 8x - 154 and find the zeros of the equation. Then

Mathematics

1 answer:

Tema [17]3 years ago

6 0

Answer:

zeros: -11 , 7

Line of symmetry: x = -2

Step-by-step explanation:

y = 2x² + 8x - 154 = 2*(x² + 4x - 77) = 2*(x+11)(x-7)

y = 0

x = - 11 or x = 7

line of symmetry: x = 1/2 (-11 + 7) = -2

2x² + 8x - 154 = 2*(x² + 4x +4) - 162 = 2 * (x+2)² - 162

vertex (h , k): h = -2 , k = -162

Line of symmetry: x = -2

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Choose the correct description of the graph of the compound inequality x − 1 less than or equal to 9 or 2x greater than or equal

DerKrebs [107]

X - 1 < = 9 or 2x > = 24
x < = 9 + 1 x > = 24/2
x < = 10 x > = 12

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closed circle on 12, shaded to the right

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3 years ago

Equation Y equals 7X can be used to find the price of different amounts of honey

Fantom [35]

Answer:

x=y/7

Step-by-step explanation:

So, the variable x is the number of jars of honey.

y is the total price of the honey.

So, x=y/7

or,

y=7x

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4 years ago

A gambler has a coin which is either fair (equal probability heads or tails) or is biased with a probability of heads equal to 0

yawa3891 [41]

Answer:

(a) 0.1719

(b) 0.3504

Step-by-step explanation:

For every coin the number of heads follows a Binomial distribution and the probability that x of the 10 times are heads is equal to:

$P(x)=\frac{n!}{x!(n-x)!}*p^x*(1-p)^{10-x}$

Where n is 10 and p is the probability to get head. it means that p is equal to 0.5 for the fair coin and 0.3 for the biased coin

So, for the fair coin, the probability that the number of heads is less than 4 is:

$P(x$

Where, for example, P(0) and P(1) are calculated as:

$P(0)=\frac{10!}{0!(10-0)!}*0.5^0*(1-0.5)^{10-0}=0.0009\\P(1)=\frac{10!}{1!(10-1)!}*0.5^1*(1-0.5)^{10-1}=0.0098$

Then, $P(x$ , so there is a probability of 0.1719 that you conclude that the coin is biased given that the coin is fair.

At the same way, for the biased coin, the probability that the number of heads is at least 4 is:

$P(x\geq4 )=P(4)+P(5)+P(6)+...+P(10)$

Where, for example, P(4) is calculated as:

$P(4)=\frac{10!}{4!(10-4)!}*0.3^4*(1-0.3)^{10-4}=0.2001$

Then, $P(x\geq4 )=0.3504$ , so there is a probability of 0.3504 that you conclude that the coin is fair given that the coin is biased.

7 0

4 years ago

Anna and Bobie joined two different swimming classes at the same time.

jeyben [28]

Answer:

It would be 10 months for the price of $320.

Step-by-step explanation:

First we need to find how much Anna's swimming charge and Bobie's charge is.

PA=70+t The price of the month equals to the starting fee and monthly charge

120=70+2t The price of 2 months times the monthly charge

50=2t

25=t is Anna's monthly charge

PB=250+t is Bobies starting price and monthly charge

264=250+2t is the month charge for Bobie after 2 months

14=2t

7=t is Bobie's monthly charge

70+25x=250+7x We need to find x which is what month will satisfy both equations

25x=180+7x Subtract 70 from both sides

18x=180 Subtract 7x from both sides

x=10 divide 18 from both sides

Now we have the months that satisfy both equations we need to find the total price that matches with it. So we plug it back into both equations.

PA=70+25x PB=250+7x

PA=70+25(10) PB=250+7(10)

PA=70+250 PB=250+70

PA=320 PB=320

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3 years ago

Read 2 more answers

The per capita growth rate of many species varies temporally for a variety of reasons, including seasonality and habitat destruc

Veronika [31]

Answer:

$n(t)=n_0e^{(1-e^{-t }-t)}$

Step-by-step explanation:

If n(t) represents the population size at time t, where n is measured in individuals and t is measured in years.

$\frac{dn}{dt}=n(e^{-t }-1), n(0)=n_o$

$\frac{dn}{n}=(e^{-t }-1)dt$

Taking the integral of both sides

$\int\frac{dn}{n}=\int(e^{-t }-1)dt\\\int\frac{dn}{n}= \int e^{-t }dt-\int1dt$

ln |n| = $-e^{-t }-t+C$

Where C is integration constant

Taking the exponential of both sides

$n=e^{(-e^{-t }-t+C)}$

$n=e^{(-e^{-t }-t)}e^C\\n=Ke^{(-e^{-t }-t)}$ whee the exponential of a constant is a constant K.

When t=0, $n(0)=n_o$

$n_0=Ke^{-1$

Therefore:

$n=n_0e^{1}e^{(-e^{-t }-t)}$

$n(t)=n_0e^{(1-e^{-t }-t)}$

7 0

3 years ago