Find a number whose product with 7 is the same as its sum with 18

So I don’t really understand this so yeh pls help *pic included* by the way this is volume of cubes and cuboids and the question

faust18 [17]

Answer:

A2: 4*4*4 = 64 cm³; 6*6*6 = 216 cm³; etc...

B2: 7*4.5*4 = 126 cm³; 10.5*6*2 = 126 cm³; etc...

Step-by-step explanation:

So in A2 you have cubes, so l=b=h, and only one has to be given, called the "length of side".

In B2 they are different, hence there are 3 rows with l, b and h. But still you multiply them l x b x h

3 0

3 years ago

A rectangle has a perimeter of 30 feet and an area of 50 square feet. What are the dimensions of the rectangle.

KiRa [710]

I believe the shorter sides are 5 feet and the longer ones are 10. This is because when you add up the two shorter sides as 10, and the other two as 20, then add them up to 30. That's the perimeter. For the area, you would do 10•5 to get 50 for the area

3 0

3 years ago

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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

Emily drew a scale drawing of a house and its lot. The back patio is 42 millimeters long in the drawing. The actual patio is 27

kirill [66]

the scale factor used in Emily scale drawing is K = 1.63 mm/m

<h3>What scale did Emily use?</h3>

We know that Emily drew a scale drawing of a house and its lot.

In the scale drawing, the back patio is 42 millimeters long, and we know that the actual patio is 27 meters long.

Then the scale relates the measures 44mm and 27 meters, such that if the scale is K, we can write:

K*27m = 44mm

Solving for K, we can write:

K = (44/27) mm/m = 1.63 mm/m

So the scale factor used in Emily's scale drawing is K = 1.63 mm/m

If you want to learn more about scale factors:

brainly.com/question/25722260

#SPJ1

5 0

2 years ago

Charlie used a regression calculator to generate the equation f(x) = –0.15x + 20.1 for the ordered pairs (2, 15), (4, 21), (6, 2

aivan3 [116]

Answer: The r<span>-value for the linear function related to the ordered pairs is very close to zero, so it is not a good representation of the data. A quadratic model would better represent the data because there is a turning point within the data set. The data increases then decreases, which is what the graph of a quadratic does. </span>

8 0

3 years ago

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