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3 years ago

You paint 1/3 of a wall in 1/4 of an hour. At that rate , how long will it take you to paint one wall

Mathematics

1 answer:

kolbaska11 [484]3 years ago

6 0

Answer:

3/4 of an hour (45 minutes)

Step-by-step explanation:

It takes 15 minutes to paint 1/3 of the wall. To paint the entire wall you would have to repeat that process 3 times, so you multiply the amount of time it took by 3.

You might be interested in

How do you graph -1/5x-1 and 4/5x-6

kakasveta [241]

Answer:

The points for the given to linear equations is (5 , - 2) and (5 , - 1)

The points is plotted on the graph shown .

Step-by-step explanation:

Given as :

The two linear equation are

y = $\dfrac{-1}{5}$ x - 1 ...........1

y = $\dfrac{4}{5}$ x - 6 ...........2

Now, Solving both the linear equations

Put the value of y from eq 2 into eq 1

I.e $\dfrac{4}{5}$ x - 6 = $\dfrac{-1}{5}$ x - 1

Or, $\dfrac{4}{5}$ x + $\dfrac{1}{5}$ x = 6 - 1

Or, $\dfrac{4 + 1}{5}$ x = 5

or, $\dfrac{5}{5}$ x = 5

∴ x = 5

Now, Put the value of x in eq 1

So, y = $\dfrac{-1}{5}$ x - 1

Or, y = $\dfrac{-1}{5}$ × 5 - 1

or, y = $\dfrac{-5}{5}$ - 1

Or, y = - 1 - 1

I.e y = -2

So, For x = 5 , y = - 2

Point is ( $x_1$ , $y_1$ ) = (5 , - 2)

Again , put the value of x in eq 2

So, y = $\dfrac{4}{5}$ x - 6

Or, y = $\dfrac{4}{5}$ × 5 - 6

Or, y = $\frac{4\times 5}{5}$ - 6

Or, y = 4 - 6

I.e y = - 2

So, For x = 5 , y = - 2

Point is ( $x_2$ , $y_2$ ) = (5 , - 2)

Hence, The points for the given to linear equations is (5 , - 2) and (5 , - 2)

The points is plotted on the graph shown . Answer

5 0

3 years ago

Whose up for mixed numbers fraction easy but have four pages

Alex777 [14]

3. A
2. C
Idk if you should repost this with a better pic though

6 0

4 years ago

Read 2 more answers

A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

What is the value of the expression k – 14 when k = –17?

Vedmedyk [2.9K]

I hope this helps you

-14+?= -17

?= -17+14

?= -3

6 0

4 years ago

Which plan to prove ∆ABD ≅ ∆CBD CANNOT be used based on the information in the diagram?

Kryger [21]

The plan that cannot be used to prove that the two triangles are congruent based in the given information is: b. ASA.

<h3>How to Prove Two Triangles are Congruent?</h3>

The following theorems can be used to prove that two triangles are congruent to each other:

SSS: This theorem proves that two triangles are congruent when there's enough information showing that they have three pairs of sides that are congruent to each other.
ASA: This theorem shows that of two corresponding angles of two triangles and a pair of included congruent sides are congruent to each other.
SAS: This theorem shows that if two triangles have two pairs of sides and a pair of included angle that are congruent, then both triangles are congruent to each other.

The two triangles only have a pair of corresponding congruent angles, while all three corresponding sides are shown to be congruent to each other.

This means that ASA which requires two pairs of congruent angles, cannot be used to prove that both triangles are congruent.

The answer is: b. ASA.

Learn more about congruent triangles on:

brainly.com/question/1675117

#SPJ1

5 0

1 year ago