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LekaFEV [45]
3 years ago
7

Sarah has gone four weeks. Micah has been recording the daily high temperature. During that time, the high temperature has been

greater than 45F on 20 out of 28 days. What is the experimental probability that the high temperature will be below 45F on the twenty-ninth day?
Mathematics
1 answer:
pickupchik [31]3 years ago
8 0
The experimental probability that the high temp will be below 45 F on the 29th day is, 4.5 out of 29 days, that is because 20 of the days have had lower temperatures, so they left 9 days of and from those 9 days, they could be lower or greater that 45F (50% each side), that is:
4.5/29 = 0.155 = 15.5%
that is the exp probability
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Karolina [17]

The line passing through (3, -1) and (-1, 5) has this slope:

5+1

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We need to find the y-intercept. To do this, substitute the knowns (-3/2 for m, -1 for y and 3 for x) into the slope-intercept equation:

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The desired equation is y = (-3/2)x + 7/2.

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3 years ago
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Tomtit [17]

Answer:

Six

Step-by-step explanation:

1.Convert all volumes to millilitres

(a) Chlorpheniramine

V = 100 mL

(b) Lidocaine

V = \text{2 fl oz} \times \dfrac{\text{29.57 mL}}{\text{1 fl oz}} = \textbf{59 mL}

(c) Banana flavouring

V = \dfrac{1}{2}\text{ tsp} \times \dfrac{\text{4.928 mL}}{\text{1 tsp}} = \textbf{2.5 mL }

2. Calculate the total volume

Chlorpheniramine = 100  mL

Lidocaine               =   59

Banana flavouring =  <u>   2.5    </u>

TOTAL                    =  162 mL

3. Calculate the number of doses

\text{Doses} = \text{162 mL} \times \dfrac{\text{1 dose}}{\text{10 mL}} = \text{16.2 doses}\\\\\text{You can't take a fraction of a dose, so the there are $\large \boxed{\textbf{16 doses}}$}

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2 years ago
Use implicit differentiation to find an equation of the tangent line to the curve at the given point. x2/3 + y2/3 = 4 (−3 3 , 1)
vovikov84 [41]

Answer with Step-by-step explanation:

We are given that an equation of curve

x^{\frac{2}{3}}+y^{\frac{2}{3}}=4

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By using implicit differentiation, differentiate w.r.t x

\frac{2}{3}x^{-\frac{1}{3}}+\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=0

Using formula :\frac{dx^n}{dx}=nx^{n-1}

\frac{2}{3}y^{-\frac{1}{3}}\frac{dy}{dx}=-\frac{2}{3}x^{-\frac{1}{3}}

\frac{dy}{dx}=\frac{-\frac{2}{3}x^{-\frac{1}{3}}}{\frac{2}{3}y^{-\frac{1}{3}}}

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Substitute the value x=-3\sqrt3,y=1

Then, we get

\frac{dy}{dx}=-\frac{(-3\sqrt3)^{-\frac{1}{3}}}{1}

\frac{dy}{dx}=-(-3^{\frac{3}{2}})^{-\frac{1}{3}}=-\frac{1}{-(3)^{\frac{3}{2}\times \frac{1}{3}}}=\frac{1}{\sqrt3}

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Equation of tangent line with slope m and passing through the point (x_1,y_1) is given by

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Substitute the values then we get

The equation of tangent line is given by

y-1=\frac{1}{\sqrt3}(x+3\sqrt3)

y-1=\frac{x}{\sqrt3}+3

y=\frac{x}{\sqrt3}+3+1

y=\frac{x}{\sqrt3}+4

This is required equation of tangent line to the given curve at given point.

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Answer:

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