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2 years ago

6

During scuba lessons, Sue dove 30 feet, Harriet dove 120 feet, and Kathy dove 90 feet.What integers represent these depths? Orde

r the integers from least to greatest. (Remember, the depth is underneath the surface of the water.)

1 answer:

Paul [167]2 years ago

4 0

-120,-90,-30. Least to greatest

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Which of the following is equivalent to 1 cup

kipiarov [429]

1 cup=16 tablespoons

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2 years ago

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Len [333]

Answer:

isp2odbskoxnxndkekjrhsbnzmcnfjhdhdhhdhdhshjsh

7 0

2 years ago

ollegr [7]

Answer:

$FC=25.9$

Step-by-step explanation:

<u>Congruent Triangles</u>

The figure shows two triangles FCD and FED. One of their sides is the radius of the circle, they both form a 90° angle with the line CD and ED and they share the same segment FD.

Thus, both triangles are congruent and we have the length of a side (12 units), one common side (the hypotenuse) and one unknown leg that must be equal. Therefore

$13x-16=4x+11$

Solving

$9x=27$

$x=3$

The required side FD is the hypotenuse of any of the triangles. The unknown leg can be calculated by

$13x-16=23$

Or

$4x+11=23$

Thus

$FD^2=12^2+23^2=673$

Solving

$FD=\sqrt{673}$

$\boxed{FC=25.9}$

5 0

2 years ago

A foreign student club lists as its members 2 Canadians, 3 Japanese, 5 Italians, and 2 Germans. If a committee of 4 is selected

Fittoniya [83]

Answer:

(a) The probability that the members of the committee are chosen from all nationalities $=\frac{4}{33}$ =0.1212.

(b)The probability that all nationalities except Italian are represent is 0.04848.

Step-by-step explanation:

Hypergeometric Distribution:

Let $x_1$ , $x_2$ , $x_3$ and $x_4$ be four given positive integers and let $x_1+x_2+x_3+x_4= N$ .

A random variable X is said to have hypergeometric distribution with parameter $x_1$ , $x_2$ , $x_3$ , $x_4$ and n.

The probability mass function

$f(x_1,x_2.x_3,x_4;a_1,a_2,a_3,a_4;N,n)=\frac{\left(\begin{array}{c}x_1\\a_1\end{array}\right)\left(\begin{array}{c}x_2\\a_2\end{array}\right) \left(\begin{array}{c}x_3\\a_3\end{array}\right) \left(\begin{array}{c}x_4\\a_4\end{array}\right) }{\left(\begin{array}{c}N\\n\end{array}\right) }$

Here $a_1+a_2+a_3+a_4=n$

${\left(\begin{array}{c}x_1\\a_1\end{array}\right)=^{x_1}C_{a_1}= \frac{x_1!}{a_1!(x_1-a_1)!}$

Given that, a foreign club is made of 2 Canadian members, 3 Japanese members, 5 Italian members and 2 Germans members.

$x_1$ =2, $x_2$ =3, $x_3$ =5 and $x_4$ =2.

A committee is made of 4 member.

N=4

(a)

We need to find out the probability that the members of the committee are chosen from all nationalities.

$a_1$ =1, $a_2$ =1, $a_3$ =1 , $a_4$ =1, n=4

The required probability is

$=\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\1\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{2\times 3\times 5\times 2}{495}$

$=\frac{4}{33}$

=0.1212

(b)

Now we find out the probability that all nationalities except Italian.

So, we need to find out,

$P(a_1=2,a_2=1,a_3=0,a_4=1)+P(a_1=1,a_2=2,a_3=0,a_4=1)+P(a_1=1,a_2=1,a_3=0,a_4=2)$

$=\frac{\left(\begin{array}{c}2\\2\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\2\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\1\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$ $+\frac{\left(\begin{array}{c}2\\1\end{array}\right)\left(\begin{array}{c}3\\1\end{array}\right) \left(\begin{array}{c}5\\0\end{array}\right) \left(\begin{array}{c}2\\2\end{array}\right) }{\left(\begin{array}{c}12\\4\end{array}\right) }$

$=\frac{1\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 2}{495}+\frac{2\times 3\times 1\times 1}{495}$

$=\frac{6+12+6}{495}$

$=\frac{8}{165}$

=0.04848

The probability that all nationalities except Italian are represent is 0.04848.

6 0

3 years ago

Micah is thinking of a 2-digit number. It is a multiple of 6 and 12. It is a factor of 108. The sum of its digits is 9. What num

Anna007 [38]

So we need to find the common multiples of 6 and 12. Which are basically the multiples of 12, since 6 is always divisible to 12. So (12, 24, 36, 48, 60, 72, 84, 96, 108) Since it is a factor of 108, we stop at 108 (since a number greater than 108 can't be a factor of 108)

Now we find the factors of 108
108=<span>1,2,3,4,6,9,12,18,27,36,54,108

The numbers in both lists are 36 and 108 but since Micah is thinking of a 2-digit number, the number she is thinking of is 36.</span>

3 0

2 years ago