SELECT ALL of the Angles that are supplementary to Angle 7<br> **NO LINKS**

Which situation can be modeled by the inequality 60−15x≥7?

Pepsi [2]

Answer:

D

Step-by-step explanation:

To find a situation for 60−15x≥7, look for a story where 60 is a constant value that is decreasing by 15 for a number of times and where equal to or at least 7 is an option.

A. x would be weeks and it would decrease by 7x. This is not it.

B. 15 songs packages would have x be the number of packages not the price. This is not it.

C. This doesn't work either because if you want to owe them less than 7 it would have the sign < not >.

D. You have 60 to spend and you decrease it by $15 for each hat you buy till you save $7. This is it.

8 0

3 years ago

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Psychologist Michael Cunningham conducted a survey of university women to see whether, upon graduation, they would prefer to mar

bagirrra123 [75]

Answer:

A.The probability that exactly six of Nate's dates are women who prefer surgeons is 0.183.

B. The probability that at least 10 of Nate's dates are women who prefer surgeons is 0.0713.

C. The expected value of X is 6.75, and the standard deviation of X is 2.17.

Step-by-step explanation:

The appropiate distribution to us in this model is the binomial distribution, as there is a sample size of n=25 "trials" with probability p=0.25 of success.

With these parameters, the probability that exactly k dates are women who prefer surgeons can be calculated as:

$P(x=k) = \dbinom{n}{k} p^{k}(1-p)^{n-k}\\\\\\P(x=k) = \dbinom{25}{k} 0.25^{k} 0.75^{25-k}\\\\\\$

A. P(x=6)

$P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.00024*0.00423=0.183\\\\\\$

B. P(x≥10)

$P(x\geq10)=1-P(x$

$P(x=0) = \dbinom{25}{0} p^{0}(1-p)^{25}=1*1*0.0008=0.0008\\\\\\P(x=1) = \dbinom{25}{1} p^{1}(1-p)^{24}=25*0.25*0.001=0.0063\\\\\\P(x=2) = \dbinom{25}{2} p^{2}(1-p)^{23}=300*0.0625*0.0013=0.0251\\\\\\P(x=3) = \dbinom{25}{3} p^{3}(1-p)^{22}=2300*0.0156*0.0018=0.0641\\\\\\P(x=4) = \dbinom{25}{4} p^{4}(1-p)^{21}=12650*0.0039*0.0024=0.1175\\\\\\P(x=5) = \dbinom{25}{5} p^{5}(1-p)^{20}=53130*0.001*0.0032=0.1645\\\\\\P(x=6) = \dbinom{25}{6} p^{6}(1-p)^{19}=177100*0.0002*0.0042=0.1828\\\\\\$

$P(x=7) = \dbinom{25}{7} p^{7}(1-p)^{18}=480700*0.000061*0.005638=0.1654\\\\\\P(x=8) = \dbinom{25}{8} p^{8}(1-p)^{17}=1081575*0.000015*0.007517=0.1241\\\\\\P(x=9) = \dbinom{25}{9} p^{9}(1-p)^{16}=2042975*0.000004*0.010023=0.0781\\\\\\$

$P(x\geq10)=1-(0.0008+0.0063+0.0251+0.0641+0.1175+0.1645+0.1828+0.1654+0.1241+0.0781)\\\\P(x\geq10)=1-0.9287=0.0713$

C. The expected value (mean) and standard deviation of this binomial distribution can be calculated as:

$E(x)=\mu=n\cdot p=25\cdot 0.25=6.25\\\\\sigma=\sqrt{np(1-p)}=\sqrt{25\cdot 0.25\cdot 0.75}=\sqrt{4.69}\approx2.17$

4 0

2 years ago

Jenna has at most $80 to spend on clothes. She wants to buy a pair of jeans for$22 and spend the rest on t-shirts. Each t-shirt

weqwewe [10]

Answer:

7 shirts

Step-by-step explanation:

You first have to subtract 22 from 80, since that's how much she already spent, 80-22=58. Now you have to do 58/8, but since 8 doesn't divide evenly into 58, you have to go with the closet multiple of 8 which would be 56. 7x8=56, so Jenna can buy a maximum number of 7 shirts.

5 0

2 years ago

Cos(x+y)/cos(x)sin(y)=cot(y)-tan(x)

zhuklara [117]

So there is an identity we'll need to use to solve this:

cos(x+y) = cosxcosy - sinxsiny

replace the numerator with the right hand side of that identity and we get:

(cosxcosy - sinxsiny)/cosxsiny

Separate the numerator into 2 fractions and we get:

cosxcosycosxsiny- sinxsiny/cosxsiny

the cosx's cancel on the left fraction, the siny's cancel on the right fraction and we're left with:

cosy/siny - sinx/cosx

which simplifies to:

coty - tanx

7 0

2 years ago

Marshall began making a grilled cheese sandwich at 6:13 P.M. The sandwich was ready at 6:24 P.M. How long did it take Marshall t

jasenka [17]

It took him 11 minutes.

3 0

2 years ago

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SELECT ALL of the Angles that are supplementary to Angle 7 **NO LINKS**

SELECT ALL of the Angles that are supplementary to Angle 7 NO LINKS