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4 years ago

Which graph correctly solves the system of equations below? y=-x^2+1 y=x^2-4

Mathematics

1 answer:

KIM [24]4 years ago

5 0

Red line shows
y=x^2-4 , cause it's going through point (0,-4)

So blue line shows
y= -x^2+1 Point (1,0) belong to it , moreover x factor is negative , so the arms of parabola will go down

Have nice evenig :)
Greetings From Poland.

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It takes 56 pounds of seed to completely plant a 7-acre field.

Xelga [282]

Answer:

there are 8 acres that can be planted

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3 years ago

Determine which of the values is a solution of 17 - 2x > 5.

Nezavi [6.7K]

Answer: x ≤ 6

Solution: -10, -4, 0, 3, 6

Not a solution: 10

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5 0

4 years ago

Read 2 more answers

Let vector F = (6 x^2 y + 2 y^3 + 4 e^x) i + (7 e^{y^2} + 54 x) j . Consider the line integral of vector F around the circle of

balu736 [363]

Denote the circle of radius $a$ by $C$ . $C$ is simple and closed, so by Green's theorem the line integral reduces to a double integral over the interior of $C$ (call it $D$ ):

$\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\int_C(6x^2y+2y^3+4e^x)\,\mathrm dx+(7e^{y^2}+54x)\,\mathrm dy$

$=\displaystyle\iint_D\left(\frac{\partial(7e^{y^2}+54x)}{\partial x}-\frac{\partial(6x^2y+2y^3+4e^x)}{\partial y}\right)\,\mathrm dx\,\mathrm dy$

$=\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy$

$D$ is a circle of radius $a$ , so we can write the double integral in polar coordinates as

$\displaystyle\iint_D(54-6x^2-6y^2)\,\mathrm dx\,\mathrm dy=\int_0^{2\pi}\int_0^a(54-6r^2)r\,\mathrm dr\,\mathrm d\theta$

a. For $a=1$ , we have

$\displaystyle\int_0^{2\pi}\int_0^1(54-6r^2)r\,\mathrm dr\,\mathrm d\theta=2\pi\int_0^1(54r-6r^3)\,\mathrm dr=\boxed{51\pi}$

b. Let $I(a)$ denote the integral with unknown parameter $a$ ,

$I(a)=12\pi\int_0^a(9r-r^3)\,\mathrm dr\,\mathrm d\theta$

By the fundamental theorem of calculus,

$I'(a)=12\pi(9a-a^3)$

$I(a)$ has critical points when

$12\pi(9a-a^3)=12\pi a(9-a^2)=0\implies a=0,a=\pm3$

If $a=0$ , then line integral is 0, so we ignore that critical point. For the other two, we would find $I(\pm3)=243\pi$ .

8 0

3 years ago

a hospital spokesperson states that 2% of emergency rooms visits by college undergraduates are for alcohol related health proble

Mazyrski [523]

Answer:

Type I error: Reject H₀: p = 0.02 when in fact p = 0.02.

Type II error: Fail to Reject H₀: p = 0.02 when in fact p ≠ 0.02.

Step-by-step explanation:

The complete question is:

Write a sentence describing the type I and type II errors for the hypothesis test for the indicated claim.

"A hospital spokesperson states that 2% of emergency rooms visits by college undergraduates are for alcohol related health problems."

Solution:

A type I error occurs when we discard a true null hypothesis (H₀) and a type II error is made when we fail to discard a false null hypothesis (H₀).

The claim made by the hospital spokesperson is, 2% of emergency rooms visits by college undergraduates are for alcohol related health problems.

The hypothesis can be defined as:

H₀: The proportion of emergency rooms visits by college undergraduates for alcohol related health problems is 2%, i.e. p = 0.02.

Hₐ: The proportion of emergency rooms visits by college undergraduates for alcohol related health problems is not 2%, i.e. p ≠ 0.02.

A type I error will be committed when we conclude that the null hypothesis can be rejected, i.e. proportion of emergency rooms visits by college undergraduates for alcohol related health problems is not 2%, when in fact we are rejecting a true null hypothesis, i.e. the proportion is 2%.

A type II error will be committed when we conclude that the null hypothesis cannot be rejected, i.e. proportion of emergency rooms visits by college undergraduates for alcohol related health problems is 2%, when in fact we are failing to reject a true null hypothesis, i.e. the proportion is not 2%.

8 0

3 years ago

9) Write a function with the following information: initial value is 300,

finlep [7]

Answer:

P(t) = 300e^0.15t

Step-by-step explanation:

The exponential equation for calculating the growth rate is expressed as;

P = P0e^rt

P0 is the initial value =300,

r is the growth rate = 15% = 0.15

t is the time

Substitute;

P = 300e^0.15t

Hence the function that represent the information is P = 300e^0.15t

7 0

3 years ago