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fenix001 [56]
4 years ago
14

Solve -3 1/2 = 1/2x + 1/2x + x

Mathematics
1 answer:
Elenna [48]4 years ago
6 0

- 3 1/2 = 2x

-7 /2   = 2 x

-7 /4  = x

-1 3/4 = x

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Find the area of the regular 18-gon with a radius of 3 mm. <br> Round to the nearest tenth.
marta [7]

Answer:

this becomes 1/2 * 6 * tan(10) * 3 which results in the area of one of the isosceles triangles is equal to 1.586942826 square units. since there are 18 of these isosceles triangles in the polygon, then multiply this by 18 to get area of the polygon with 18 sides is equal to 28.56497087 square units.

Follow me please

Mark brainliest

7 0
3 years ago
Based on data from a statistical abstract, only about 16% of senior citizens (65 years old or older) get the flu each year. Howe
SSSSS [86.1K]

Answer: The required probability is 2.24%.

Step-by-step explanation:

Since we have given that

Percentage of people are senior citizens ( 65 years old or older ) = 14%

Percentage of people are under 65 years old = 100-14 = 86%

Probability that senior citizens get the flu each year = 16%

Probability that under 65 years old get the flu each year = 30%

So, Probability that a person selected at random from the general population is senior citizen who get the flu this season is given by

P(S\cap F)={P(F|S)\times P(S)\\\\P(S\cap F)=0.16\times 0.14\\\\P(S\cap F)=0.0224\\\\P(S\cap F)=2.24\%Hence, the required probability is 2.24%.

3 0
3 years ago
Read 2 more answers
Statistics Probability
andriy [413]

Answer:

Explained below.

Step-by-step explanation:

Denote the variable as follows:

M = male student

F = female student

Y = ate breakfast

N = did not ate breakfast

(a)

Compute the probability that a randomly selected student ate breakfast as follows:

P(Y)=\frac{n(Y)}{N}\\\\=\frac{198}{330}\\\\=0.60

(b)

Compute the probability that a randomly selected student is female and ate breakfast as follows:

P(F\cap Y)=\frac{n(F\cap Y)}{N}\\\\=\frac{121}{330}\\\\=0.367

(c)

Compute the probability a randomly selected student is male, given that the student ate breakfast as follows:

P(M|Y)=\frac{n(M\cap Y)}{n(Y)}\\\\=\frac{77}{198}\\\\=0.389

(d)

Compute the probability that a randomly selected student ate breakfast, given that the student is male as follows:

P(Y|M)=\frac{n(Y\cap M)}{n(M)}\\\\=\frac{77}{154}\\\\=0.50

(e)

Compute probability of the student selected "is male" or "did not eat breakfast" as follows:

P(M\cup N)=P(M)+P(N)-P(M\cap N)\\\\=\frac{n(M)}{N}+\frac{n(N)}{N}-\frac{n(M\cap N)}{N}\\\\=\frac{n(M)-n(N)-n(M\cap N)}{N}\\\\=\frac{154+132-77}{330}\\\\=0.633

(f)

Compute the probability of "is male and did not eat breakfast as follows:

P(M\cap N)=\frac{n(M\cap N)}{N}\\\\=\frac{77}{330}\\\\=0.233

4 0
3 years ago
Who's recipe will have a stronger strawberry flavor? show you work and/or explain your reasoning
Alexxandr [17]

Answer:

what

Step-by-step explanation:

where is the problem bro

3 0
3 years ago
Given that it was less than 80º on a given day, what is the probability that it also rained that day? 0.3 0.35 0.65 0.7
eimsori [14]

The probability that it also rained that day is to be considered as the 0.30 and the same is to be considered.

<h3>What is probability?</h3>

The extent to which an event is likely to occur, measured by the ratio of the favorable cases to the whole number of cases possible.

The probability that the temperature is lower than 80°F and it rained can be measured by determining the number at the intersection of a temperature that less than 80°F and rain.

So, This number is 0.30.

Hence, we can say that it was less than 80°F on a given day, the probability that it also rained that day is 0.30.

To learn more about the probability from the given link:

brainly.com/question/18638636

The above question is incomplete.

The conditional relative frequency table was generated using data that compared the outside temperature each day to whether it rained that day. A 4-column table with 3 rows titled weather. The first column has no label with entries 80 degrees F, less than 80 degrees F, total. The second column is labeled rain with entries 0.35, 0.3, nearly equal to 0.33. The third column is labeled no rain with entries 0.65, 0.7, nearly equal to 0.67. The fourth column is labeled total with entries 1.0, 1.0, 1.0. Given that it was less than 80 degrees F on a given day, what is the probability that it also rained that day?

#SPJ4

7 0
2 years ago
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