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3 years ago

If the triangle shown is translated vertically 3 units and horizontally -4 units graph the image of the triangle

1 answer:

7 0

Okay so you need to write down all of your original points, for example:
<A (1,1)
<B (3,4)
<C (5,1)
so now for each of those coordinates you would subtract four from each X and add 3 to each Y. For example:
Original Cord.:                             Translated Cord.:
<A (1,1)                                       <A' (-3,4)
<B (3,4)                                       <B' (-1,6)
<C (5,1)                                       <C' (2, 4)
------------------------------------------------------------------------
So just follow my steps but with your points, then graph both the original coordinates and the translated coordinates. AND DON'T FORGET TO LABEL THE POINTS LIKE I DID WHEN GRAPHING!!!!!!!!

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What is the value of 5 to the power of 4 over 5 to the power of 6?

belka [17]

5^4÷ 5^6=5^-2

1 over 25

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3 years ago

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The inequality 2x+1>15 will be true when<br> 1. x=7<br> 2. x=0<br> 3. x=10<br> 4. x=5

pogonyaev

Answer:

3.) x = 10

Step-by-step explanation:

Let’s plug it into our equation

2(10) + 1 = 21

21 is greater than 15

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2 years ago

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.

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

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3 years ago

Suppose a die is tossed 5 times. what is the probability of getting exactly 2 fours?

Tasya [4]

Answer:

the probability of getting exactly 2 fours is 0.16

Step-by-step explanation:

The probability of obtaining a number that is four = ¹/₆

The probability of obtaining a non 4 number = 1 - ¹/₆ = ⁵/₆

The number of ways 2 fours can be arrange in five numbers = ⁵C₂ = 10 ways

If the die is tossed five times, the probability of the events is calculated as;

P = 10 x (¹/₆)² x (⁵/₆)³

P = 10 x (¹/₃₆) x (¹²⁵/₂₁₆)

P = 10 x 0.02778 x 0.5787

P = 0.16

Therefore, the probability of getting exactly 2 fours is 0.16

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2 years ago

Of the students in Ms.Danver's class, 6 walk to school.This represents 30% of her students.How many Students are in Ms.Damver's

Kisachek [45]

There is 20 students. To find the answer you need to use porportionals and cross multiply.

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3 years ago

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