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Lerok [7]
3 years ago
6

Assume the probability of having a boy or a girl is the same. What is the probability a family has three boys?

Mathematics
1 answer:
almond37 [142]3 years ago
3 0
The probability of having a boy: 1/2

1/2(1/2)(1/2)=1/8

The probability of a family having three boys is 1/8.

Hope this helps!
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Can you help me find each measure?
madam [21]

Answer:

Step-by-step explanation:

1. the measure for HJ you can already see, start from h and draw a line to J. The measure shows 63 degrees.

2. start at F and draw a line to G first. the measure shows 65 degrees. Now continue the line from where you stopped at G, to H. There's no measure but you can see it is in a semicircle. A semicircle is 180 degrees andthe other two angles are 63 and 65.

So...

180=63+65+GH

subtract to get GH

180-63-65= 52

so the measure GH is 52 degrees. The full measure you are trying to find is FGH thought. So ad the 65 and the 52.

FGH =117 degrees.

3. The meaure is CDE. CD as you can see is a right angles, so 90 degrees. But there is no measure for DE. If you look to the angle vertical from DE which is BA. It measures 40 degrees. DE and BA are vertical so they are congruent. If DE equals 40  and CD equals 90, put them together and you get 130.

CDE= 130 degrees

4. Next measure is BCD. We already know CD is a 90 degree angle but BC is blank. You can see the measure BCD is in a semi circle. A semicircle equals 180, CD equals 90, and BA equals 40

so...

180= 90+40+BC

so subtract

180-90-40= 50

BC =50

so add BC=50 and CD=90. SO, BCD is 140 degrees.

5.The angle LMN is next. MN is 30 but LM is blank. LMN is in a semi circle.

A semicircle is 180 degrees.

so...

180=105+30+LM

subtract

180-105-30=45

LM = 45

Add LM=45 and MN=30

LMN is 75 degrees.

6.The last angle is LNP

If you look at it MP is a semi circle, so 180 degrees. And LM is 45 from our last question. so 180 +45=225

LNP=225

hope this helps

4 0
3 years ago
Can a sequence be both arithmetic and geometric?
artcher [175]
An aritmetic sequence is like this
a_n=a_1+d(n-1) where a1=first term and d=common difference

geometric is a_n=a_1(r)^{n-1} where a1=first term and r=common ratio


can it be both aritmetic and geometric
hmm, that means that the starting terms should be the same

therfor we need to solve d(n-1)=(r)^{n-1}
what values of d and r make all natural numbers of n true?
are there values that make all natural numbers for n true?

when n=1, then d(1-1)=0 and r^(1-1)=1, so already they are not equal

the answer is no, a sequence cannot be both aritmetic and geometric
4 0
3 years ago
Solve using substitution<br><br> y=1/2x<br> x+4y=18
Aleks [24]

Answer:

The solution set is (6,3)

Step-by-step explanation:

Begin by multiplying equation 1 by 2. You'll see why in a moment

2y = 2*(1/2) x

2y = x

Substitute this value into equation 2.

2y + 4y = 18                  Combine like terms on the left

6y = 18                          Divide both sides by 6

6y/6 = 18/6                    Do the division

y = 3

To solve for x just use the top equation

y = 1/2x

3 = 1/2x                          Multiply by 2

3*2 = x

x = 6


7 0
3 years ago
Can any one help me solve this ?
Lana71 [14]

Step 1. Take out the constants

(4 * 7)xx^2

Step 2. Simplify 4 * 7 to 28

28xx^2

Step 3. Use the Product Rule

28x^1 + 2

Step 4. Simplify 1 + 2 to 3

28x^3

3 0
3 years ago
I have honestly no clue how to do this plz help me
Crank

Answer:

M_{2} = \frac{r^{2} F_{g} }{GM_{1}}

Step-by-step explanation:

\frac{GM_{1} M_{2}}{r^{2}}=F_{g}

Multiply both sides by (r²)....  GM_{1} M_{2}=r^{2} F_{g}

Divide both sides by (GM₁).... M_{2} = \frac{r^{2} F_{g} }{GM_{1}}

6 0
3 years ago
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