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pav-90 [236]
3 years ago
15

Can someone tell me what I did wrong.

Mathematics
2 answers:
finlep [7]3 years ago
5 0

Answer:

when adding fractions make sure the bottom number is the same

so for number 2 make sure both are 4 on the bottom of the fraction by multiplying by 2 maybe 1/4+2/4 that would equal 3/4 not 1

nignag [31]3 years ago
3 0

Answer:

Step-by-step explanation:

1/2+1/4=0.75 less than 1

3/6+2/3=1.17 greater than 1

2/4+1/3=10/12 less than 1

mark brainliest thx

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Question is 15 points
Blababa [14]

sin(x+y)=sin(x)cos(y)-cos(x)sin(y)

also, remember pythagorean rule, sin^2(x)+cos^2(x)=1

given that sin(Θ)=4/5 and cos(x)=-5/13

find sin(x) and cos(Θ)


sin(x)

cos(x)=-5/13

using pythagorean identity

(sin(x))^2+(-5/13)^2=1

sin(x)=+/- 12/13

in the 2nd quadrant, sin is positve so sin(x)=12/13


cos(Θ)

sin(Θ)=4/5

using pythagrean identity

(4/5)^2+(cos(Θ))^2=1

cos(Θ)=+/-3/5

in 1st quadrant, cos is positive

cos(Θ)=3/5


so sin(Θ+x)=sin(Θ)cos(x)+cos(Θ)sin(x)

sin(Θ+x)=(4/5)(-5/13)+(3/5)(12/13)

sin(Θ+x)=16/65



answer is 1st option

7 0
3 years ago
A) Find a​ 90% confidence interval for the slope of the regression line of​ %Body Fat on Waist.
dusya [7]

Answer:

this is to hard and I like math

4 0
3 years ago
Find the probability of being dealt a full house. (Round your answer to six decimal places.)?
vesna_86 [32]

I'm assuming a 5-card hand being dealt from a standard 52-card deck, and that there are no wild cards.

A full house is made up of a 3-of-a-kind and a 2-pair, both of different values since a 5-of-a-kind is impossible without wild cards.

Suppose we fix both card values, say aces and 2s. We get a full house if we are dealt 2 aces and 3 2s, or 3 aces and 2 2s.

The number of ways of drawing 2 aces and 3 2s is

\dbinom42\dbinom43=24

and the number of ways of drawing 3 aces and 2 2s is the same,

\dbinom43\dbinom42=24

so that for any two card values involved, there are 2*24 = 48 ways of getting a full house.

Now, count how many ways there are of doing this for any two choices of card value. Of 13 possible values, we are picking 2, so the total number of ways of getting a full house for any 2 values is

2\dbinom{13}2\dbinom42\dbinom43=3744

The total number of hands that can be drawn is

\dbinom{52}5=2,598,960

Then the probability of getting a full house is

\dfrac{2\binom{13}2\binom42\binom43}{\binom{52}5}=\dfrac6{4165}\approx\boxed{0.001441}

4 0
3 years ago
A football player completes 19 of his 25 passes during the season. What percent of his passes did the player complete? Use fract
d1i1m1o1n [39]
19/25 is the answer for this one.
5 0
3 years ago
A(n) __________ is a conic section formed when a plane intersects one nappe of a double-napped cone, parallel to the generating
allochka39001 [22]

Answer:

it's parabola

Step-by-step explanation:

don't worry I did my research on this one:)

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