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3 years ago

A couple plans to have 3 children, what is the probability that atleast two will be boys?

2 answers:

6 0

1/4 or 25 percent because there is a 1/2 chance to get a boy or girl so the probability that will be boys us 1/2 x 1/2= 1/4

givi [52]3 years ago

6 0

Answer:

<h3>There are 2^3=8 possible combinations for two genders for three children.</h3><h3 /><h3>There are 2^3=8 possible combinations for two genders for three children.There are (GGG, GGB, BGG, GBG) = 4 combinations where at least two of the children are girls.</h3><h3 /><h3>There are 2^3=8 possible combinations for two genders for three children.There are (GGG, GGB, BGG, GBG) = 4 combinations where at least two of the children are girls.Therefore, there is a 4/8 = 1/2 probability that there will be at least two girls in the family of three children.</h3><h3 />

Step-by-step explanation:

<h2><u>PLEASE</u><u> </u><u>MARK</u><u> ME</u><u> BRAINLIEST</u><u> AND</u><u> FOLLOW</u><u> ME</u><u> LOTS</u><u> OF</u><u> LOVE</u><u> FROM</u><u> MY</u><u> HEART'AND</u><u> SOUL</u><u> DARLING</u><u> TEJASVINI</u><u> SINHA</u><u> HERE</u><u> ❤️</u><u>❤️</u><u>❤️</u><u /></h2>

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The County Fair has to 2 ticket options.Option 1 has an entry fee of $5 and charges $0.65 per ride per ride. Option 2 has an ent

Ugo [173]

For 25 tickets total cost of option one and option two to be the same.

Step-by-step explanation:

Let us assume the total number of rides = m

for which BOTH options cost same.

Case: 1

The cost of entry fee = $5

The cost per ride = $0.65

So, the cost of m rides = m x ( Cost of 1 ride )

= m x ($0.65 ) = 0.65 m

Cost of purchasing m tickets in first ride = Entry Fee + Per ticket cost

= 5 +0.65 m ..... (1)

Case: 2

The cost of entry fee = $10

The cost per ride = $0.45

So, the cost of m rides = m x ( Cost of 1 ride )

= m x ($0.45 ) = 0.45 m

Cost of purchasing m tickets in second ride = Entry Fee + Per ticket cost

= 10 +0.45 m ..... (2)

Now, equating (1) and (2), we get:

5 +0.65 m = 10 +0.45 m

or, 0.20 m = 5

or, m = 5/0.20 = 25

or, m = 25

Hence, for 25 tickets total cost of option one and option two to be the same.

5 0

3 years ago

The ratio of boys to girls in the ninth grade is 6:5. If there are 185 girls in the class, how many boys are there?

xeze [42]

Answer:

$222$ boys

Step-by-step explanation:

Let $x$ = total number of ninth graders. By knowing how many students there are in the grade, we can find for the number of boys.

$\frac{5}{6+5}x = 185$ (number of girls in ninth grade)

$\frac{5}{11}x = 185$

$x=407$

To find for the number of boys, we can use substitution:

$\frac{6}{6+5}x = \frac{6}{11}(407)$

$=222$

∴ There are $222$ boys in the class.

Hope this helps :)

8 0

2 years ago

Read 2 more answers

The daily cost of producing x high performance wheels for racing is given by the following function, where no more than 100 whe

WINSTONCH [101]

So all you need to do if find the min of the function:

f(x)=0.04x^3-4x^2-176x
f'(x)=0.12x^2-8x-176
The only zero for this is x=84.105 <---first
Plug this into f(x) to get f(84.105)=-19300 <---second

4 0

3 years ago

What is the multiplicative inverse of 5 in z11, z12, and z13? you can do a trial-and-error search using a calculator or a pc?

grin007 [14]

A multiplicative inverse of an integer a is an integer x such that the product ax is congruent to 1 with respect to the modulus m.

1. $Z_{11}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10}\}.$

Check:

$5\cdot 0=\overline{0};$
$5\cdot 1=\overline{5};$
$5\cdot 2=\overline{10};$
$5\cdot 3=15=\overline{4};$
$5\cdot 4=20=\overline{9};$
$5\cdot 5=25=\overline{3};$
$5\cdot 6=30=\overline{8};$
$5\cdot 7=35=\overline{2};$
$5\cdot 8=40=\overline{7};$
$5\cdot 9=45=\overline{1};$
$5\cdot 10=50=\overline{6}.$

The multiplicative inverse of 5 in $Z_{11}$ is 9.

2. $Z_{12}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11}\}.$

Check:

$5\cdot 0=\overline{0};$
$5\cdot 1=\overline{5};$
$5\cdot 2=\overline{10};$
$5\cdot 3=15=\overline{3};$
$5\cdot 4=20=\overline{8};$
$5\cdot 5=25=\overline{1};$
$5\cdot 6=30=\overline{6};$
$5\cdot 7=35=\overline{11};$
$5\cdot 8=40=\overline{4};$
$5\cdot 9=45=\overline{9};$
$5\cdot 10=50=\overline{2};$
$5\cdot 11=55=\overline{7}.$

The multiplicative inverse of 5 in $Z_{12}$ is 5.

3. $Z_{13}=\{\overline{0},\overline{1},\overline{2},\overline{3},\overline{4},\overline{5},\overline{6},\overline{7},\overline{8},\overline{9},\overline{10},\overline{11},\overline{12}\}.$