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kondaur [170]
3 years ago
13

Determine the value of w and x in the figure below. (Thanks for the help I really appreciate it!)​

Mathematics
1 answer:
storchak [24]3 years ago
7 0
Triangle HAM looks like an isosceles triangle, and we know that in an isosceles triangle 2 of the angles are the same.

so :
180 - 106 = 74 ( sum of the other 2 angles )
74/2 = 37 ( sum of angle HMA / HAM )

angles on a straight line add up to 180.
180-37=143 ( angle x )

since triangle YMH is also an isosceles,
180-37=143
143/2=71.5 ( angle w )
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Which values of x satisfy this equation √x^2-10x+25+12√x=15√x
Effectus [21]

Answer:

If I am not mistaken, I believe the answers are 1.425 and 17.57, so A and D.

Step-by-step explanation:

5 0
3 years ago
If the length of the base of a triangle is
Anarel [89]

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d

Step-by-step explanation:

the answer is 12% decrease in area

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2 years ago
Solve 2x^2 + 3 = 5 by factoring. (use the big “X” method please!)
Talja [164]

Answer: x = -1, 1

Step-by-step explanation:

7 0
2 years ago
Of the entering class at a​ college, ​% attended public high​ school, ​% attended private high​ school, and ​% were home schoole
Veronika [31]

Answer:

(a) The probability that the student made the​ Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

<em>A</em> = a student attended public high school

<em>B</em> = a student attended private high school

<em>C</em> = a student was home schooled

<em>D</em> = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})

Compute the probability that the student made the​ Dean's list as follows:

P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)

         =(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655

Thus, the probability that the student made the​ Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

P(B|D)=\frac{P(D|B)P(B)}{P(D)}

             =\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

P(C^{c}|D^{c})=1-P(C|D^{c})

               =1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0
2 years ago
Please help me. these problems<br>​
jeyben [28]

Answer:

1st problem:

Converges to 6

2nd problem:

Converges to 504

Step-by-step explanation:

You are comparing to \sum_{k=1}^{\infty} a_1(r)^{k-1}

You want the ratio r to be between -1 and 1.

Both of these problem are so that means they both have a sum and the series converges to that sum.

The formula for computing a geometric series in our form is \frac{a_1}{1-r} where a_1 is the first term.

The first term of your first series is 3 so your answer will be given by:

\frac{a_1}{1-r}=\frac{3}{1-\frac{1}{2}}=\frac{3}{\frac{1}{2}=6

The second series has r=1/6 and a_1=420 giving me:

\frac{420}{1-\frac{1}{6}}=\frac{420}{\frac{5}{6}}=420(\frac{6}{5})=504.

3 0
2 years ago
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