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Schach [20]
4 years ago
14

The measure of ∠1 = 188 172 94

Mathematics
1 answer:
yanalaym [24]4 years ago
4 0

Answer:  Third option is correct.

Step-by-step explanation:

Since we have given that

There is a cyclic quadrilateral.

As we know that "Sum of opposite angles in a cyclic quadrilateral is supplementary.":

so, it becomes,

86^\circ+\angle 1=180^\circ\\\\\angle 1=180^\circ-86^\circ\\\\\angle 1=94^\circ

Hence, the measure of  ∠1 = 94°

Therefore, Third option is correct.

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A cake has a circumference of 25 1/7 inches. What is the area of the cake? Use 22/7 to approximate pi
Novosadov [1.4K]
To calculate for area given the circumference we proceed as follows:
C=2πr
where:
r=radius
thus given that C=25 1/7 in=176/7
thus plugging in our formula to solve for r we get:
176/7=2πr
thus
r=4.002~4.00 in
hence the area will be:
A=πr²
A=π×(4²)
A=50 2/7 in²


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4 years ago
There are 20 squares and 12 triangles. What is the simplest ratio of triangles to total shapes?
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3:8 is the answer ….
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3 years ago
Record your<br> ! 1)<br> 5x = 125
KengaRu [80]
The answer is X= 25
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3 0
3 years ago
4x−2y=14 y=12x−1 solve by substitution
Tatiana [17]

Answer:

x= -3/5

Step-by-step explanation:

4x-2y=14

y=12x-1

4x-2(12x-1)=14

4x-24x+2=14

4x-24x=12

-20x=12

×= -12/20

x= -3/5

8 0
3 years ago
Samples of 20 parts from a metal punching process are selected every hour. Typically, 1% of the parts require rework. Let X deno
Roman55 [17]

Answer:

a) P(X>np+3\sqrt{np(1-p)}=0.017

b) P(x>1)=0.190

c) P(Y>1)=0.651

Step-by-step explanation:

This a binomial experiment where success is denoted by parts that need rework.

X ∼ B(n, p); n = 20; p = 0.01

The expected value of X is: E(X) = np =20×0.01= 0.2

The variance is: Var(X) = np(1 − p) = 0.2 × 0.99 = 0.198,

The standard deviation SD(X)= \sqrt{0.198} ≈ 0.445

a) P(X>np+3\sqrt{np(1-p)}=P(X>0.2+3×0.445)=P(X>1.535)=P(X≥2)

Probability function is given by:

\frac{n!}{x!(n-x)!} *p^x*(1-p)^{(n-x)}

P(X≥2)=1-P(X<2)=1-P(X=1)-P(X=0)= 1 - \frac{20!}{1!(20-1)!} *(0.01)^{1}*(1-0.01)^{(20-1)}-\frac{20!}{0!(20-0)!} *(0.01)^{0}*(1-0.01)^{(20-0)}

P(X≥2)=1-0.165-0.818=0.017

b) p=0.04

P(x>1)=P(x≥2)= 1 - P(x=1) - P(x=0)= 1 - \frac{20!}{0!(20-1)!} *(0.04)^{1}*(1-0.04)^{(20-1)} - \frac{20!}{0!(20-0)!} *(0.04)^{0}*(1-0.04)^{(20-0)}

P(x>1)= 1 - 0.368 - 0.442=0.190

c) In this case we consider p=0.19 (Probability that X exceeds 1)

In this experiment Y is the number of hours and n= 5 hours.

Then, we check the probability in each hour:

P(Y>1)=1- P(Y=0)

P(Y=0)=\frac{5!}{0!(5-0)!} *(0.19)^{0}*(1-0.19)^{(5-0)}=0.349

P(Y>1)=1-0.349=0.651

3 0
3 years ago
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