Which expression is equivalent to 8 square root 6 ?

A class has a ratio of boys to girls 3:4.

Nat2105 [25]

Answer:

a)F

b)A

c)F

d)F

Step-by-step explanation:

a)3+4=7 which is the total number of pupils in class

So, the fraction should be 3/7✅not 3/4❌

b) The girls in the class are 4 and the total number of pupils in class are 7. So, the fraction is correct✅.

c) The number of boys in class is 3✅ which is given in the ratio. Not 6❌

d) If you add the both numbers in the ratio, you will get the total number of pupils which is 7✅not 12❌

Hope this will help ya!

4 0

2 years ago

Read 2 more answers

Pls help and find the area

Zepler [3.9K]

Answer:

I don’t know the area but here’s a hint to help you (hint: length x with x height)

Step-by-step explanation:

5 0

2 years ago

What is the simplified form of each expression?<br>1.7m^2 + 6.5n - 4n + 2.5m^2 - n

Ulleksa [173]

${4.2m}^{2} + 1.5n$

3 0

2 years ago

Simplify. (x−3)( x^2−4x−7)

Shkiper50 [21]

Answer:

Simplification of the expression (x−3)( x^2−4x−7) is x³ -7x² + 5x + 21 .

Step-by-step explanation:

As the expression given in the question be as follow .

= (x−3)( x²−4x−7)

simplify the above

= x (x²-4x-7)-3 (x²-4x-7)

Now open the bracket

= x³ - 4x² -7x - 3x² + 12x + 21

= x³- 4x²-3x²-7x + 12x + 21

= x³ -7x² + 5x + 21

Thus the simplification of the expression (x−3)( x^2−4x−7) is x³ -7x² + 5x + 21 .

5 0

2 years ago

Read 2 more answers

A data set includes 40 pulse rates of men, and those pulse rates have a mean of 67.3 beats per minute and a standard deviation o

Dafna11 [192]

Answer:

The 99% confidence interval of the standard deviation of the pulse rates of men is

$[ 7.949 \ \ , \ 14.387 ]$

Step-by-step explanation:

From the question we are told that

The sample size is n = 40

The mean is $\= x = 67.3 \ beats \ per \ minute$

The standard deviation is $s = 10.3 \ beats \ per \ minute$

Generally the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 40- 1$

=> $df = 39$

From the question we are told the confidence level is 99% , hence the level of significance is

$\alpha = (100 - 99 ) \%$

=> $\alpha = 0.01$

Generally from the chi distribution table the critical value of at a degree of freedom of is

$X^2 _{\frac{\alpha }{2} , 39} = 65.476$

Generally from the chi distribution table the critical value of at a degree of freedom of is

$X^2 _{1- \frac{\alpha }{2} , 39} = 19.996$

Generally the 99% confidence interval of the standard deviation of the pulse rates of men is mathematically represented as

$s \sqrt{\frac{n- 1}{X^2_{1- \frac{\alpha }{2} , n-1} } }\ \ , \ s \sqrt{\frac{n- 1}{X^2_{\frac{\alpha }{2} , n-1} } }$

=> $10.3 \sqrt{\frac{40- 1}{65.476} }\ \ , \ 10.3 \sqrt{\frac{40- 1}{19.996} }$

=> $[ 7.949 \ \ , \ 14.387 ]$

6 0

2 years ago