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3 years ago

If m B = 3x - 9 and m C = 39°, then what is the value of x? A.10 B.44 C.47 D.50

Mathematics

1 answer:

GalinKa [24]3 years ago

4 0

I think its 44
If I'm wrong pleas tell me
but I hope I'm right

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student tickets for a football game cost $2 each. adult tickets cost $4 each. ticket sales at last week's game totaled $2800. wr

ELEN [110]

Answer:

Let s = students

Let a = adults

2s + 4a = 2800

Step-by-step explanation:

6 0

3 years ago

Jose, baulus, and George are having a disagreement over interest rates. Jose says that 6 3/4% can be expressed as 6.75%. George

otez555 [7]

Answer:

Jose and George are correct

Baulus is incorrect

Step-by-step explanation:

Given

$6\frac{3}{4}$ %

Required

Compare the above expression with

Jose: 6.75%.

George: 0.0675

Baulus: 0.0675%

<em>The expression </em> $6\frac{3}{4}$ <em>% is given as a mixed fraction;</em>

In Jose's case, he simply converted it to decimal

Converting $6\frac{3}{4}$ % to decimal;

First, we need to convert to improper fraction

$6\frac{3}{4}$ % in improper fraction is $\frac{27}{4}$ %

Then, we divide the numerator by the denominator

$\frac{27}{4}$ % = 6.75%.

Hence, Jose is correct

In George's case, he solved further by removing the % from Jose's answer.

% means divided by 100; In other words, x% means that we divide x by 100

So, 6.75% means we divide 6.75 by 100

$\frac{6.75}{100}$ = 0.0675

Hence, George is also correct

But same can't be said of Baulus;

Baulus converted $6\frac{3}{4}$ % as 0.0675%

This means that he converted $6\frac{3}{4}$ as 0.0675 then he appended it with the percentage sign (%)

Since $6\frac{3}{4}$ is not equal to 0.0675, then we can conclude that Baulus is incorrect

8 0

4 years ago

-(j-7)+3=5 what is j?

Fofino [41]

The answer is 5/ j=5

7 0

3 years ago

A survey report states that 70% of adult women visit their doctors for a physical examination at least once in two years. If 20

irakobra [83]

Answer:

a) 0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

b) 0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

Step-by-step explanation:

For each women, there are only two possible outcomes. Either they visit their doctors for a physical examination at least once in two years, or they do not. The probability of a woman visiting their doctor at least once in this period is independent of any other women. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

70% of adult women visit their doctors for a physical examination at least once in two years.

This means that $p = 0.7$

20 adult women

This means that $n = 20$

(a) Fewer than 14 of them have had a physical examination in the past two years.

This is:

$P(X < 14) = 1 - P(X \geq 14)$

In which

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 14) = C_{20,14}.(0.7)^{14}.(0.3)^{6} = 0.1916$

$P(X = 15) = C_{20,15}.(0.7)^{15}.(0.3)^{5} = 0.1789$

$P(X = 16) = C_{20,16}.(0.7)^{16}.(0.3)^{4} = 0.1304$

$P(X = 17) = C_{20,14}.(0.7)^{17}.(0.3)^{3} = 0.0716$

$P(X = 18) = C_{20,18}.(0.7)^{18}.(0.3)^{2} = 0.0278$

$P(X = 19) = C_{20,19}.(0.7)^{19}.(0.3)^{1} = 0.0068$

$P(X = 20) = C_{20,20}.(0.7)^{20}.(0.3)^{0} = 0.0008$

$P(X \geq 14) = P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.1916 + 0.1789 + 0.1304 + 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.6079$

$P(X < 14) = 1 - P(X \geq 14) = 1 - 0.6079 = 0.3921$

0.3921 = 39.21% probability that fewer than 14 of them have had a physical examination in the past two years.

(b) At least 17 of them have had a physical examination in the past two years

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)$

From the values found in item (a).

$P(X \geq 17) = P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20) = 0.0716 + 0.0278 + 0.0068 + 0.0008 = 0.107$

0.107 = 10.7% probability that at least 17 of them have had a physical examination in the past two years.

6 0

3 years ago

Which of the following are solutions to x-1 = 5x + 2 ? Check all that apply.

Ad libitum [116K]

Answer:

The answer is-3/4

Step-by-step explanation:

#Hope it helps uh.....

5 0

3 years ago