Two angles supplementary to the same angle must be congruent.

Find the outer perimeter of this figure.

stira [4]

Answer:

61.90

Step-by-step explanation:

12.2+14+14+6=46.2

of semi circle

circumference of semi circle = $\pi$ d

$\pi$ x10 = 10 $\pi$

semi circle so divide by 2 = 5 $\pi$

5 $\pi$ = 15.7

46.2 + 15.7 = 61.9

3 0

2 years ago

What do you know about the slope when X2 - X1 = 0 ?

Viktor [21]

Answer:

slope is undefined

Step-by-step explanation:

Using the slope formula

m = (y₂ - y₁ ) / (x₂ - x₁ )

If x₂ - x₁ = 0

Since division by zero is undefined then the slope of the line is undefined

This applies to a vertical line parallel to the y- axis

4 0

3 years ago

PLEASE HELP Solve the following systems of inequalities and select the correct graph:

NNADVOKAT [17]

Answer:

que duro a fuego hermano

Step-by-step explanation:

3 0

3 years ago

Read 2 more answers

A rectangular banner covers an area of 2 7/8ft2. The length of the banner is 3/4ft. What is the width of the banner in ft?

hichkok12 [17]

Answer:

Width of the rectangular banner is $3\frac{5}{6}$ ft.

Step-by-step explanation:

Area of the rectangular banner = $2\frac{7}{8}$ square feet

= $\frac{23}{8}$ square feet

Area of rectangle is given by,

Area = Length × Width

Length of the banner = $\frac{3}{4}$ feet

From the formula,

$\frac{23}{8}=\frac{3}{4}\times W$

W = $\frac{\frac{23}{8} }{\frac{3}{4} }$

= $\frac{23}{8}\times \frac{4}{3}$

= $\frac{23}{6}$

= $3\frac{5}{6}$ ft

Therefore, width of the rectangular banner is $3\frac{5}{6}$ ft

7 0

3 years ago

There is a 70 percent chance that an airline passenger will check bags. In the next 16 passengers that check in for their flight

GenaCL600 [577]

Answer:

a) $P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332$

b) $P(X$

And we can find the individual probabilities like this:

$P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649$

$P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099$

$P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040$

$P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465$

$P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732$

$P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228$

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033$

And replacing we got:

$P(X$

c) $P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)$

And replacing we got 0.825

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=17, p=0.7)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

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

Part a

And we want to find this probability:

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.00332$

Part b fewer than 10 will check bags

We want this probability:

$P(X$

We can use the complement rule and we have:

$P(X$

And we can find the individual probabilities like this:

$P(X=10)=(16C10)(0.7)^{10} (1-0.7)^{16-10}=0.1649$

$P(X=11)=(16C11)(0.7)^{11} (1-0.7)^{16-11}=0.2099$

$P(X=12)=(16C12)(0.7)^{12} (1-0.7)^{16-12}=0.2040$

$P(X=13)=(16C13)(0.7)^{13} (1-0.7)^{16-13}=0.1465$

$P(X=14)=(16C14)(0.7)^{14} (1-0.7)^{16-14}=0.0732$

$P(X=15)=(16C15)(0.7)^{15} (1-0.7)^{16-15}=0.0228$

$P(X=16)=(16C16)(0.7)^{16} (1-0.7)^{16-16}=0.0033$

And replacing we got:

$P(X$

Part c at least 10 bags

We can find this probability like this:

$P(x \geq 10) = P(X=10)+P(X=11)+P(X=12)+P(X=13)+P(X=14)+P(X=15)+P(X=16)$

And replacing we got 0.825

4 0

3 years ago