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lukranit [14]
4 years ago
10

Pls help me I'm taking a test and it's due in 30 mins idk this.

Mathematics
2 answers:
padilas [110]4 years ago
8 0

Answer:

36-12y=(x-2)^2 or y=-(x^2/12) + x/3+8/3

Step-by-step explanation:

inessss [21]4 years ago
4 0

Answer:

. y2 = 12x 40:12 :3 2. x² = -8y 408 3. (x + 2)2 = -4(y-1). \(-2,1) opens v. F-2,0) ... F (0:2). V(0,-2). F(4,2)

Step-by-step explanation:

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The body temperatures of a group of healthy adults have a​ bell-shaped distribution with a mean of 98.18 F and a standard deviat
S_A_V [24]

Answer:

99.7%

Step-by-step explanation:

Empirical rule formula states that:

• 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ.

• 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ.

• 99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ.

From the question, we have mean of 98.18 F and a standard deviation of 0.65 F

The approximate percentage of healthy adults with body temperatures between 96.23 F and100.13 ​F is

μ - 3σ

= 98.18 - 3(0.65)

= 98.18 - 1.95

= 96.23 F

μ + 3σ.

98.18 + 3(0.65)

= 98.18 + 1.95

= 100.13 F

Therefore, the approximate percentage of healthy adults with body temperatures between 96.23 F and 100.13 ​F which is within 3 standard deviations of the mean is 99.7%

6 0
4 years ago
For what value of constant c is the function k(x) continuous at x = 0 if k =
nlexa [21]

The value of constant c for which the function k(x) is continuous is zero.

<h3>What is the limit of a function?</h3>

The limit of a function at a point k in its field is the value that the function approaches as its parameter approaches k.

To determine the value of constant c for which the function of k(x)  is continuous, we take the limit of the parameter as follows:

\mathbf{ \lim_{x \to 0^-} k(x) =  \lim_{x \to 0^+} k(x) =  0 }

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= c }

Provided that:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}= \dfrac{0}{0} \ (form) }

Using l'Hospital's rule:

\mathbf{\implies  \lim_{x \to 0} \ \  \dfrac{\dfrac{d}{dx}(sec \ x - 1)}{\dfrac{d}{dx}(x)}=  \lim_{x \to 0}   sec \ x  \ tan \ x = 0}

Therefore:

\mathbf{\implies  \lim_{x \to 0 } \ \  \dfrac{sec \ x - 1}{x}=0 }

Hence; c = 0

Learn more about the limit of a function x here:

brainly.com/question/8131777

#SPJ1

5 0
2 years ago
For the function y=-3+4 cos(5pi/6(x+4)) what is the minimum value
Vikentia [17]
Interval <-1;1>, which means that this function can take values from interval <-7;-1>. The minimum value is -7.
5 0
3 years ago
Read 2 more answers
Plz someone answer these 2 questions.
Elan Coil [88]

Answer:

First one: 0.4

Second one: 4

Step-by-step explanation:

We find the cube root of 0.064 which is 0.4

In order to find the length of side s, we just cube root 64 because it is s^{3}

3 0
3 years ago
Read 2 more answers
Leroy's total job benefits were $50,150 last year. His total job expenses for traveling and for professional development for the
Diano4ka-milaya [45]

Answer:

D. 53,650

50,150+3,500=53,650

7 0
3 years ago
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