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3 years ago

Conrad has 11 containers for his toys. He put 17 toys in each container. How many toys did he put away in all? *

Mathematics

1 answer:

ArbitrLikvidat [17]3 years ago

6 0

Answer:

A child's toy consists of a piece of plastic attached to a spring in the figure below. The spring is compressed against the floor a distance of 2.00 cm, and the toy is released. If the toy has a mass of 117 g and rises to a maximum height of 45.0 cm, estimate the force constant of the spring.

Step-by-step explanation:

You might be interested in

Find the smallest power of 10 that will exceed M. Let M = 118,526.65902.

zavuch27 [327]

Answer:

The smallest power of 10 that will exceed $M$ is $10^{6}$ .

Step-by-step explanation:

We can use the following approach to determine the smallest power of 10 that will exceed M. We can transform that number into scientific notation, which is of the form:

$x.y \times 10^{n}$ , $\forall \,x,y\in \mathbb{N}$

Where:

$x$ - Integer part, formed by a digit, which is of the highest order.

$y$ - Decimal part, formed by a digit onwards.

$n$ - Power grade.

The smallest power of 10 that will exceed M is $10^{n+1}$

If $M = 118,526.65902$ , then, the power grade is number of spaces that dot must be moved leftwards. In this case, dot must be moved 5 spaces on the left. The integer part is 1 and the decimal part is 1852665902. Then, the value of $M$ in scientific notation is:

$M = 1.1852665902\times 10^{5}$

Then, the smallest power of 10 that will exceed $M$ is $10^{6}$ .

5 0

3 years ago

Please help, I beg of y'all <br><br>FIND BD -

puteri [66]

9514 1404 393

Answer:

14 units

Step-by-step explanation:

The angle bisector divides the sides proportionally.

BD/BA = CD/CA

(x+4)/8 = (2x+1)/12

3(x +4) = 2(2x +1) . . . . . multiply by 24 (the least common denominator)

3x +12 = 4x +2 . . . . . . . eliminate parentheses using the distributive property

12 = x +2 . . . . . . . subtract 3x

10 = x . . . . . . . . . subtract 2

Then the length of BD is ...

BD = x +4

BD = 10 +4 . . . . substitute the value of x

BD = 14

3 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

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

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

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

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

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

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

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

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

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

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