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3 years ago

Justin bought 6 ribbons for an art project. Each ribbon is 1/4 yard long. How many yards of ribbon did Justin buy?

2 answers:

8 0

To answer this question, Justin has bought 3/2 yards of the ribbon.
There are two ways you could answer this question. You could add it or subtract it. First, I will teach you the way of adding to get your answer. So, from this sentence, Justin has bought 6 ribbon and each was 1/4 yards long. You could add it by doing 1/4+1/4+1/4+1/4+1/4+1/4. This would give you 6/4, which would be 3/2 yards(simplified). (This could also be 1.5 yards..)
The second way was multiplying. It would look like this: 1/4 times 6. (Now, 6, could be represent as 6/1) Now you do 1 time 6 for your numerator and 1 times 4, which would be your denominator. This would also give you 6/4 or 3/2 (simplified) Hope this works!! :)

Gemiola [76]3 years ago

5 0

6 ribbons
1/4 yard long each
4 yards = 4 ribbons
4/4 yards
A: 6/4 yd

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Graph the solution for the following linear inequality system. Click on the graph until the final result is displayed.

-Dominant- [34]

Answer:

y ≥ 0 Pic 1

y < x pic 2

x + y < 6 pic 3

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3 years ago

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For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger

11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let X = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = $p=\frac{1}{120}$ .

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...$

In this case we need to compute the probability of 1 or more than 1 catastrophes in n missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}$

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166$

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410$

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803$

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419$

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3 years ago

Look at the graph below. what is the slope of the line ?

kykrilka [37]

Answer:

The slope is $-\frac{1}{5}$ .

Step-by-step explanation:

Choose 2 points on the graph.

(7,3) and (2,2)

Use slope formula: $\frac{y2-y1}{x2-x1}$

$\frac{2-3}{2-7}$ = $\frac{-1}{-5}$

The slope is $-\frac{1}{5}$ .

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2 years ago

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A taxi service offers a ride with an $5 surcharge and charges $0.50 per mile. How many miles can a customer travel and spend at

hram777 [196]

Charges 0.50 per mile with a 5 dollar surcharge. Spending at most 30 dollars, the equation would be:

0.5x+5 $\leq$ 30

To find how many miles they can travel/the solution, solve the inequality:

$0.5x+5\leq30$

Subtract 5 from both sides.

$0.5x\leq25$

Divide both sides by 0.5.

$x\leq50$

They can travel (at most) 50 miles.

Hope this helps :)

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3 years ago

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Solve:

Advocard [28]

First we look at the equation an determine our common denominator.

(X²-X-6) / X² = (X-6) / 2X + (2X+12) / X (let's use 2X² as the comm denom)

multiply the first fraction by 2, the second by X, and the third by 2x

(2x²-2x-12) / 2x² = (x²-6x) / 2x² + (4x²+24x) / 2x² (dropping the denominator)

2x² - 2x - 12 = x² - 6x + 4x² + 24x (simplify)

-3x² - 20x - 12 (multiply by -1)
3x² + 20x + 12
(x + 6) (3x + 2) (factored version)

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3 years ago

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