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3 years ago

A die is rolled. What is the probability of rolling the following?

Mathematics

2 answers:

Anton [14]3 years ago

7 0

$|\Omega|=6\\|A|=3+1=4\\\\P(A)=\dfrac{4}{6}=\dfrac{2}{3}\approx67\%$

Rus_ich [418]3 years ago

6 0

Answer:

2/3 chance (67%)

Step-by-step explanation:

Multiples of 5:

(5)

Multiples of 2:

(2)

(4)

(6)

This is a 4/6 chance or 2/3

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What is the area of the parallelogram?

Anarel [89]

Answer:

7 and one-half feet squared

Step-by-step explanation:

3 x 2.5 = 7.5

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2 years ago

A store is having a sale on almonds and jelly beans. For 2 pounds of almonds and 5 pounds of jelly beans, the total cost is $17

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I think it would be $16 :)

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3 years ago

What is the future value for a savings account that had $825 deposited in it for 10 years. The

Goshia [24]

Answer:

The answer is 2145$

Step-by-step explanation:

825×0.04=33

Once every quarter of a year means the number of years × 4 = 10×4=40

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2 years ago

Use the parabola tool to graph the quadratic function y=-1? - 2.0 +8

Andreas93 [3]

Answer:

see explanation

Step-by-step explanation:

I don't have graphing facilities but can give you the vertex and 1 other point.

Given a parabola in standard form

y = ax² + bx + c ( a ≠ 0 )

Then the x- coordinate of the vertex is

x = - $\frac{b}{2a}$

y = - x² - 2x + 8 ← is in standard form

with a = - 1 and b = - 2 , then

x = - $\frac{-2}{-2}$ = - 1

Substitute x = - 1 into the equation for corresponding value of y

y = - (- 1)² - 2(- 1) + 8 = - 1 + 2 + 8 = 9

vertex = (- 1, 9 )

To obtain another point substitute any value for x into the equation

x = 0 : y = 0 - 0 + 8 , then (0, 8 ) is a point on the graph

x = 2 : y = - (2)² - 2(2) + 8 = - 4 - 4 + 8 = 0 then (2, 0 ) is a point on the graph

3 0

3 years ago

You are working in a small, student-run company that sends out merchandise with university branding to alumni around the world.

topjm [15]

Answer:

The answer is "0.142466".

Step-by-step explanation:

Using the p formula for the proportion of nonconforming units through the subgrouping which can vary in sizes:

$p =\frac{np}{n}\\\\$

$\bar{p}=\frac{\Sigma np}{\Sigma n}\\\\$

Defects $= \frac{5}{100} \times 50 \\\\$

$p = \frac{5}{100} \times \frac{50}{50}=0.05\\\\$

It calculates the controls limits through the p-chart that is:

$UCL_{p},LCL_{p}=\bar{p} \pm \sqrt{\frac{\bar{p}(1-\bar{p})}{\bar{n}}}\\\\$

So, the upper control limits:

$= 0.05 + 3 \times \sqrt{(\frac{0.05\times(1-0.05)}{50})} \\\\= 0.142466$

7 0

2 years ago