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3 years ago

Five year old Michelle weighs 75 decagrams. What is her weight in centigrams?

Mathematics

2 answers:

kvv77 [185]3 years ago

5 0

75,000 should be your answer!

Svet_ta [14]3 years ago

3 0

It should be 75000. I don’t know. I looked this up but I still hope this helps!

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HELP HELP HELP HELP WOPFOPOPMPOMGOPRDAGS

Basile [38]

Answer:

Step-by-step explanation:

You can't ever let this go negative. At least not at the grade you are in.

It can be 0.

So the domain must start at x = 7

sqrt(5*7 - 35) = sqrt(0) = 0

x can have any value (including 7) between 7 and infinity. If you choose a number less than 7 (like 6) the square root will go negative and that's not to be done.

So the interval is

7 ≤ x < ∞

4 0

3 years ago

Last January, the average cost of a one-night stay at a hotel was $132.52. Since then, the hotel has changed ownership, and

Nikolay [14]

Answer:

$114.32

Step-by-step explanation:

Original Price: $132.52

increased: $34.65

Loyalty Club: -$57.80

Phone Fee: $4.95

You add all of them together, except for the Loyalty Club Fee. You subtract that fee by the total.

4 0

3 years ago

2 times the square root of 27 minus the square root of 48

lyudmila [28]

$2 \sqrt{27}- \sqrt{48}=2 \sqrt{9*3}- \sqrt{16*3} =2*3 \sqrt{3}- 4 \sqrt{3}=6 \sqrt{3}-4 \sqrt{3} = \\ =2 \sqrt{3}$

5 0

3 years ago

A ball is thrown with an initial velocity of 70 km per second, at an angle of 35° with the horizontal. Find the vertical and ho

oksian1 [2.3K]

Answer:

i. The horizontal component of the ball = 15.8

ii. The vertical component of the ball = 11.2

Step-by-step explanation:

A thrown ball is an example of a projectile, whose motion can be explained with respect to both horizontal and vertical components.

Horizontal component of the ball = U Cos θ

Vertical component of the ball = U Sin θ

where U is the initial velocity of the ball, and θ is the angle of projection.

Thus for the given question,

U = 70 km/hr (19.44 m/s)

θ = 35°

So that;

i. The horizontal component of the ball = 19.44 Cos 35°

= 19.44 x 0.8192

= 15.785

The horizontal component of the ball = 15.8.

ii. The vertical component of the ball = 19.44 Sin 35°

= 19.44 x 0.5736

= 11.151

The vertical component of the ball = 11.2

5 0

3 years ago

Of the entering class at a college, % attended public high school, % attended private high school, and % were home schoole

Veronika [31]

Answer:

(a) The probability that the student made the Dean's list is 0.1655.

(b) The probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c) The probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

Step-by-step explanation:

The complete question is:

Of the entering class at a college, 71% attended public high school, 21% attended private high school, and 8% were home schooled. Of those who attended public high school, 16% made the Dean's list, 19% of those who attended private high school made the Dean's list, and 15% of those who were home schooled made the Dean's list.

a) Find the probability that the student made the Dean's list.

b) Find the probability that the student came from a private high school, given that the student made the Dean's list.

c) Find the probability that the student was not home schooled, given that the student did not make the Dean's list.

Solution:

Denote the events as follows:

A = a student attended public high school

B = a student attended private high school

C = a student was home schooled

D = a student made the Dean's list

The provided information is as follows:

P (A) = 0.71

P (B) = 0.21

P (C) = 0.08

P (D|A) = 0.16

P (D|B) = 0.19

P (D|C) = 0.15

(a)

The law of total probability states that:

$P(X)=\sum\limits_{i} P(X|Y_{i})\cdot P(Y_{i})$

Compute the probability that the student made the Dean's list as follows:

$P(D)=P(D|A)P(A)+P(D|B)P(B)+P(D|C)P(C)$

$=(0.16\times 0.71)+(0.19\times 0.21)+(0.15\times 0.08)\\=0.1136+0.0399+0.012\\=0.1655$

Thus, the probability that the student made the Dean's list is 0.1655.

(b)

Compute the probability that the student came from a private high school, given that the student made the Dean's list as follows:

$P(B|D)=\frac{P(D|B)P(B)}{P(D)}$

$=\frac{0.21\times 0.19}{0.1655}\\\\=0.2410876\\\\\approx 0.2411$

Thus, the probability that the student came from a private high school, given that the student made the Dean's list is 0.2411.

(c)

Compute the probability that the student was not home schooled, given that the student did not make the Dean's list as follows:

$P(C^{c}|D^{c})=1-P(C|D^{c})$

$=1-\frac{P(D^{c}|C)P(C)}{P(D^{c})}\\\\=1-\frac{(1-P(D|C))\times P(C)}{1-P(D)}\\\\=1-\frac{(1-0.15)\times 0.08}{(1-0.1655)}\\\\=1-0.0815\\\\=0.9185$

Thus, the probability that the student was not home schooled, given that the student did not make the Dean's list is 0.9185.

3 0

3 years ago