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Black_prince [1.1K]
3 years ago
5

Joe owns a stock which has probability .5 of going up. This morning, he bought a ticket in a lottery game which gives him a prob

ability .0001 of winning. What is the probability that Joe's stock will go up and he will win in the lottery? Your answer should keep 5 positions to the right of the decimal, like .43212
Mathematics
1 answer:
yKpoI14uk [10]3 years ago
7 0

Answer:

0.00005

Step-by-step explanation:

The probability of the stock going up and that of winning in the lottery are independent.

Hence, the probability of both = P(stock going up) × P(winning in the lottery) = 0.5 × 0.0001 = 0.00005

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The design on Santana’s wall includes 16 pink stripes and 20 green stripes. Find the ratio of pink stripes to green stripes.
tia_tia [17]
The ratio is 4:5
Divide both terms by 4
16/4 =4
20/4=5

8 0
3 years ago
Help me solve this image please!!!
tresset_1 [31]

Based on the information, Christian would have $5525.5 of an annuity.

<h3>How to calculate the annuity?</h3>

According to the given information, the number of coffees per week is 3 then, per month is 3x4 = 12

Each coffee is $4.5. Then monthly expenditure for coffees is 12 x 4.5 = $54

Rate of interest r = 1.6% = 1.6/100 = 0.016 and for monthly compounding r = 0.016/12 = 0.00133

n = number of payments = 8 x 12 = 96

We can use the formula for finding the future value as below

FV = C x [ ( 1 + r )n-1 ] / ( r )

FV = 54 x [ ( 1 + 0.00133 )96 – 1 ] / (0.00133)

= 54 x [ (1.13609 - 1)] / (0.00133)

= 54 x 0.13609 / (0.00133)

= 54 x 102.3233

= 5525.5

Therefore Christian would have $5525.5 of the annuity.

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brainly.com/question/5303391

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5 0
2 years ago
Assume the population of human body temperatures has a mean of 98.6 degrees Fahrenheit as is commonly believed. Assume the stand
Dennis_Churaev [7]

The probability that the mean of a sample of 106 randomly selected humans is lower than 98.5°F is 4.85%

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Z score is given as:

z = (raw score - mean) ÷ (standard deviation/√sample size)

Given mean of 98.6°F,  standard deviation is 0.62°F, sample size = 106

For x < 98.5:

z = (98.5 - 98.6) ÷ (0.62÷√106) = -1.66

P(z < -1.66) = 0.0485

The probability that the mean of a sample of 106 randomly selected humans is lower than 98.5°F is 4.85%

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4 0
2 years ago
-8x-10y=24;6x+-10x-10y
stiks02 [169]
Do you have a typo? ;  < this maybe
4 0
4 years ago
Which is the equation of a hyperbola centered at the origin with x-intercept +\- 3 and asymptote y=2x
Radda [10]

Answer:

{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}

Step-by-step explanation:

The hyperbola has x-intercepts, so it has a horizontal transverse axis.

The standard form of the equation of a hyperbola with a horizontal transverse axis is  \dfrac{(x - h)^{2}}{a^{2}} - \dfrac{(y - k)^{2}}{b^{2}} = 1

The center is at (h,k).

The distance between the vertices is 2a.

The equations of the asymptotes arey = k \pm \dfrac{b}{a}(x - h)

1. Calculate h and k. The hyperbola is symmetric about the origin, so  

h = 0 and k = 0

2. For 'a': 2a = x₂ - x₁ = 3 - (-3) = 3 + 3 = 6

a = 6/2 = 3  

3. For 'b': The equation for the asymptote with the positive slope is  

y = k + \dfrac{b}{a}(x - h) = \dfrac{b}{a}x

Thus,  asymptote has the slope of

\begin{array}{rcl}m& =& \dfrac{b}{a}\\\\2& =& \dfrac{b}{3}\\\\b& =& \mathbf{6}\end{array}

4.  The equation of the hyperbola is

\large \boxed{\mathbf{\dfrac{x^{2}}{9} - \dfrac{y^{2}}{36} = 1}}

The attachment below represents your hyperbola with x-intercepts at ±3 and asymptotes with slope ±2.

7 0
3 years ago
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