Through:(1,7), slope=3

Ok so here is another question. I love math don’t y’all? (I absolutely hate it)

AlexFokin [52]

Answer:

2.5 x 12 = 30

3.5 x 2.5 = 8.75

30 + 8.75 = 38.75 ft²

5 0

3 years ago

Two linear equations are shown.

Arte-miy333 [17]

Answer:

x = 7; y = 4 1/3

Step-by-step explanation:

The solution is the point of intersection of the lines.

Read the point of intersection.

You can also solve the system using algebra.

y = 1/3 x + 2

y = 4/3 x - 5

1/3 x + 2 = 4/3 x - 5

x + 6 = 4x - 15

-3x = -21

x = 7

y = 1/3 x + 2

y = 1/3 × 7 + 2

y = 2 1/3 + 2

y = 4 1/3

Answer: x = 7; y = 4 1/3

6 0

2 years ago

Fernando has two equally-sized containers of clay which he is using to build a pyramid-like structure. The finished structure wi

evablogger [386]

Answer:

Fernando does not have enough clay to finish the structure since, the expressions for the volume left are not the same.

Step-by-step explanation:

Let w be the length of the longest slab. Since each slab is 2 inches less than the one beneath it, we have the width of the other 3 slabs as w - 2, w - 2 - 2 = w - 4 and w - 4 - 2 = w - 6 respectively.

Since the base of each slab is a square and has thickness 2 inches, the volume of each slab from largest to smallest is thus 2w², 2(w - 2)², 2(w - 4)² and 2(w - 6)² respectively.

We have that we have half of the container of clay left after forming the first two slabs(which are the bottom-most slabs). Let V be the volume of each clay container. Since we have half left, that is V/2, we have used 2V - V/2 = 3V/2 to make the two bottom-most slabs.

So, 3V/2 = 2w² + 2(w - 2)²

= 2w² + 2(w² - 4w + 4)

= 2w² + 2w² - 8w + 8

= 4w² - 8w + 8

= 4(w² - 2w + 2) (1)

Also we want to know if V/2 is equal to the volume of the two top-most slabs.

So, V/2 = 2(w - 4)² + 2(w - 6)²

= 2(w² - 8w + 16) + 2(w² - 12w + 36)

= 2w² - 16w + 32 + 2w² - 24w + 72

= 4w² - 40w + 104

= 4(w² - 10w + 26)

From (1) V/2 = 4(w² - 2w + 2)/3 ≠ 4(w² - 10w + 26)

<u>Since the expressions for the remaining volume are not the same, Fernando does not have enough clay to finish the structure.</u>

4 0

3 years ago

Ted is constructing an equilateral triangle. He draws a segment with his straightedge and labels the endpoints A and B. He then

gladu [14]

Answer:

Step-by-step explanation:

Keep the same compass opening, place the compass point on B, and draw a circle.

3 0

3 years ago

Recent homebuyers from a local developer allege that 30% of the houses this developer constructs have some major defect that wil

umka21 [38]

Answer:

3.54% probability of observing at most two defective homes out of a random sample of 20

Step-by-step explanation:

For each house that this developer constructs, there are only two possible outcomes. Either there are some major defect that will require substantial repairs, or there is not. The probability of a house having some major defect that will require substantial repairs is independent of other houses. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

30% of the houses this developer constructs have some major defect that will require substantial repairs.

This means that $p = 0.3$

If the allegation is correct, what is the probability of observing at most two defective homes out of a random sample of 20

This is $P(X \leq 2)$ when n = 20. So

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{20,0}.(0.3)^{0}.(0.7)^{20} = 0.0008$

$P(X = 1) = C_{20,1}.(0.3)^{1}.(0.7)^{19} = 0.0068$

$P(X = 2) = C_{20,2}.(0.3)^{2}.(0.7)^{18} = 0.0278$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0008 + 0.0068 + 0.0278 = 0.0354$

3.54% probability of observing at most two defective homes out of a random sample of 20

5 0

3 years ago