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3 years ago

The ratio of boys to girls in a class is 5 to 7. If there are 35 girls, how many boys are there?

Mathematics

1 answer:

-BARSIC- [3]3 years ago

5 0

Answer:

25, you have to set up a ratio between 5 to 7 and 35.

Step-by-step explanation:

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How many solutions can be found for the linear equation? 4(x + 5) - 5 = 8x + 18/ 2

mylen [45]

Answer: The correct option is A.

Explanation:

The given equation is,

$4(x+5)-5=\frac{8x+18}{2}$

Multiply both sides by 2.

$2[4(x+5)-5]=\frac{8x+18}{2}$

$8(x+5)+2(-5)=8x+18$

$8x+40-10=8x+18$

$8x+30=8x+18$

Subtract both sides by 8x.

$30=18$

This statement is false for any value of x, therefore the system of equation have no solution and option A is correct.

6 0

4 years ago

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IF A=[2,4] and B=[0,3] then BUA=?

Semenov [28]

Answer:

it equal:

answer is : [0234]

6 0

3 years ago

The scores of individual students on the American College Testing (ACT) Program College Entrance Exam have a normal distribution

tekilochka [14]

Answer:

The probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

Step-by-step explanation:

Xi~N(18.6, 6.0), n=400, Yi~Ber(p); Z~N(0, 1);

$P(0\leq X\leq 19.0)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{19-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}, \mu=18.6, \sigma=6.0$

$P(-3.1\leq Z\leq 0.0667)=\Phi(0.0667)-\Phi (-3.1)=\Phi(0.0667)-(1-\Phi (3.1))=0.52790+0.99903-1=0.52693$

P(Xi≥19.0)=0.473

$\{Yi=0, Xi< 19\\Yi=1, Xi\geq 19\}$

p=0.473

Yi~Ber(0.473)

$P(\frac{1}{n}\displaystyle\sum_{i=1}^{n}X_i\geq 19)=P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{n}X_i\~{}N(n\mu, \sqrt{n}\sigma),\displaystyle\sum_{i=1}^{400}X_i\~{}N(7440, 372)$

Then:

$P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)=1-P(0$

$P(\displaystyle\sum_{i=1}^{n}Y_i=1)=P(\displaystyle\sum_{i=1}^{400}Y_i=1)$

Based on the Central Limit Theorem:

$\displaystyle\sum_{i=1}^{400}Y_i\~{}N(400\times 0.473, \sqrt{400}\times 0.499)=\displaystyle\sum_{i=1}^{400}Y_i\~{}N(189.2; 9.98)$

$P(\displaystyle\sum_{i=1}^{400}Y_i=1)\~{=}P(0.5$

Then:

the probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0

7 0

3 years ago

Solve for z Z - 4/9 - 1/3 = 5/9

Aleks04 [339]

Answer:

4/3

Step-by-step explanation:

Z-4/9-1/3=5/9

Z-7/9=5/9

Z=5/9+7/9 

Z=4/3 not sure 

3 0

3 years ago

If a = 7 and b = 2, find the value of a) a- b b) 2a + 3b c) 2ab d) (a - b)?

Dimas [21]

A) 7-2=5
b) 2(7)+3(2)=14+6=20
c) 2(7)(2)=28
d) (7-2)=5

5 0

3 years ago

Read 2 more answers