Help im sister needs help and i can't help her im doing my school work !

Use the Pythagorean Theorem to find the distance of the missing side. (I’m sorry I need help with the whole thing)

Maurinko [17]

Answer:

1: 5.83

2: 4.12

3: 3.61

4: 5

5: 6.32

6: 5.10

Step-by-step explanation:

The Pythagorean theorem is a^2 + b^2 = c^2

For all of these you can count the number of blocks over to find two sides of the triangle.

For example: the first one has two sides, one is 3 units, the other is 5. To find the missing angle you square 3 and 5, then add that together. That is equal to 34. 34 = c^2. To get the c by itself, you then take the square root of 34. So the missing side (c) is equal to 5.83.

8 0

2 years ago

Which measures of central tendency would be affected if the outlier 11 was added to the following set 27,20,34,37,21,42,39

MrRa [10]

Answer:

Median will not be affected by the outlier.

Step-by-step explanation:

With the outlier, the mean will be dragged way down. The median will likely be about the same. Mean is non-resistant to outliers, median is resistant.

Hope this helps!

5 0

3 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.

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

5 0

3 years ago

Help me solve for z, step by step if you can

zheka24 [161]

Answer:

z=6

Step-by-step explanation:

z/3 + 1/2 = 5/2

Subtract 1/2 from each side

z/3 + 1/2-1/2 = 5/2-1/2

z/3 = 4/2

z/3 =2

Multiply each side by 3

z/3*3 = 2*3

z = 6

5 0

3 years ago

Brainliest for whoever if they get this CORRECT.

Sergio039 [100]

c

All the information you need is in the quadratic trinomial ... Read from right to left:

Find factors of 49 which ADD (+) to give 14.

These will be 7 and 7 because: 7 × 7 = 49 and 7 + 7 = 14

The + sign shows that the signs in the brackets will be the SAME, the minus (-) shows that they will be negative:

( x − 7 ) ( x − 7 ) = ( x − 7 ) 2

8 0

3 years ago