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4 years ago

Quick Help please? :D

Mathematics

1 answer:

Vikentia [17]4 years ago

4 0

I would say B; trapezoid. That is what it appears as.

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A manufacturing firm just received a shipment of 20 assembly parts, of slightly varied sizes, from a vendor. The manager knows t

Marat540 [252]

Answer:

a. 0.75

b. 0.5526

c. 0.1974

Step-by-step explanation:

Probability is typically expressed as a fraction where the numerator is the number of desired outcomes and the denominator is the total number of outcomes. In this case the suitable parts is the desired outcome and the total shipment is the total parts:

$\frac{suitable}{total}=\frac{15}{20}=\frac{3}{4}=0.75$

The probability the both the first and second part are suitable:

$\frac{15}{20}*\frac{14}{19}=\frac{210}{380}=0.5526$

Since the probability of getting a suitable part the first time is 15/20, then there is one less suitable part and one less part overall to choose on the second time, or 14/19.

Lastly, the probability of choosing a suitable part the first time is again 15/20, but the second part not being suitable would be 5/19:

4 0

3 years ago

Using the pattern, give the coefficients of (x + y)^5 and (x + y)^6

musickatia [10]

Answer:

$(x+y)^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$

$(x+y)^6=x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$

Step-by-step explanation:

In order to find the values of $(x+y)^5$ and $(x+y)^6$ , you need to apply the binomial theorem (high-level math you most likely don't need to worry about, it's easier than multiplying all the binomials together).

$(x+y)^5$ = $\sum _{i=0}^5\binom{5}{i}x^{\left(5-i\right)}y^i$ = $\frac{5!}{0!\left(5-0\right)!}x^5y^0+\frac{5!}{1!\left(5-1\right)!}x^4y^1+\frac{5!}{2!\left(5-2\right)!}x^3y^2+\frac{5!}{3!\left(5-3\right)!}x^2y^3+\frac{5!}{4!\left(5-4\right)!}x^1y^4+\frac{5!}{5!\left(5-5\right)!}x^0y^5$ = $x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5$ .

$(x+y)^6$ = $\sum _{i=0}^6\binom{6}{i}x^{\left(6-i\right)}y^i$ = $\frac{6!}{0!\left(6-0\right)!}x^6y^0+\frac{6!}{1!\left(6-1\right)!}x^5y^1+\frac{6!}{2!\left(6-2\right)!}x^4y^2+\frac{6!}{3!\left(6-3\right)!}x^3y^3+\frac{6!}{4!\left(6-4\right)!}x^2y^4+\frac{6!}{5!\left(6-5\right)!}x^1y^5+\frac{6!}{6!\left(6-6\right)!}x^0y^6$ = $x^6+6x^5y+15x^4y^2+20x^3y^3+15x^2y^4+6xy^5+y^6$ .

5 0

3 years ago

The iPhone 6 costs $350 today. If it value depreciates at the rate of 30% per year when will it be

DIA [1.3K]

And its 6 600 and 202 and add 100 which means it will be 12 THATS your answer

5 0

3 years ago

Henri built a lantern in the shape of a square pyramid. He wants to cover the entire pyramid with fabric. How much fabric will h

aliina [53]

Answer:

The answer is 44.84. I dont actually see the correct answer here.

Step-by-step explanation:

The formula for the surface area of a square pyramid is a√(a2 + 4h2). In your pyramid a=4, and h=3. This gives 44.84 as the answer.

4 0

3 years ago

Read 2 more answers

Need some help with this question please.

Ksivusya [100]

Use Pythagorean theorem
a^2 + b^2 = c^2
Plug in the info
b = 53, a = 41
41^2 + 53^2 = c^2
1681 + 2809 = c^2
c^2 = 4490
Squareroot of 4490 = 67.007
Perimeter = adding all sides
53m + 41m + 67.01m = 161.01 m

Solution: 161.01 meters

8 0

3 years ago