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2 years ago

1 by root 50 simplest rationalising factor

Mathematics

1 answer:

givi [52]2 years ago

3 0

Answer:

1/5V2 or

V2/10

Step-by-step explanation:

1/V50=1/V5^2*2=1/5V2

rationalizing by multiply by V2

1*V2/5V2*V2=V2/5*2=V2/10

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2.4 Backgammon. Backgammon is a board game for two players in which the playing pieces are moved according to the roll of two di

muminat

Answer:

The claim is not fair because:

Probability of roliing two 6s = Probability of rolling two 3s = 1/36

Step-by-step explanation:

The probaility of an event is defined as the number of favorable outcomes divided by the number of total possible events.

P (event E) = number of outcomes for event E / number of possible events

1. As first step, you may draw a table to find the sample space (set of all possible outcomes).

Sample space

The results of the rolling two dice are summarized in this table:

Second roll 1 2 3 4 5 6

First roll

1 (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

2 (2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

3 (3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

4 (4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

5 (5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

6 (6,1) (6,2) (6,3) (6,4) (6,5) (6,6)

2.Now, in that table, you can observe:

The results (3,3) and (6,6) are highlited.

a) Total number of events: 6 × 6 = 36

b) Nnmber of outcomes for the event rolling two 6s (6,6): 1

c) Number of outcomes for the event rolling two 3s (3,3): 1

3) Next, you can calculate the probabilities:

a) Probability rolling two 6s = 1 / 36

b) Probability of rolling two 3s: 1 / 36

4. Conclusion: the probabilities prove that rolling two 6s is just as likely as rolling two 3s.

7 0

2 years ago

An ostrich can run 6 mph faster than a giraffe. An ostrich can run 7 miles in the same time that a giraffe can run 6 miles. Find

LUCKY_DIMON [66]

NOTE THIS IS AN EXAMPLE:

Let t = time, s = ostrich, and g = giraffe.

Here's what we know:

s = g + 5 (an ostrich is 5 mph faster than a giraffe)

st = 7 (in a certain amount of time, an ostrich runs 7 miles)

gt = 6 (in the same time, a giraffe runs 6 miles)

We have a value for s, so plug it into the first equation:

(g + 5)t = 7

gt = 6

Isolate g so that we can plug that variable value into the equation:

g = 6/t

so that gives us:

(6/t + 5)t = 7

Distribute:

6 + 5t = 7

Subtract 6:

5t = 1

Divide by 5:

t = 1/5 of an hour (or 12 minutes)

Now that we have a value for time, we can plug them into our equations:

1/5 g = 6

multiply by 5:

g = 30 mph

s = 30 + 5

s = 35 mph

Check by imputing into the second equation:

st = 7

35 * 1/5 = 7

7 = 7

5 0

2 years ago

Initially, there were only 197 weeds at a park. The weeds grew at a rate of 25% each week. The following function represents the

Anarel [89]

Answer:

D.) f(x) = 197(1.03)^(7x); grows at a rate of approximately 3% daily

Step-by-step explanation:

The growth equation can be written in terms of a rate compounded 7 times per week:

f(x) = 197×1.25^x = 197×(1.25^(1/7))^(7x)

f(x) ≈ 197×1.0324^(7x) . . . . x represents weeks, a daily growth factor is shown

The daily growth rate as a percentage is the difference between the daily growth factor and 1, expressed as a percentage:

(1.0324 -1) × 100% = 3.24%

The best match is choice D:

f(x) ≈ 197(1.03^(7x)); grows approximately 3% daily

4 0

2 years ago

Read 2 more answers

Data from 14 cities were combined for a 20-year period, and the total 280 city-years included a total of 107 homicides. After

leonid [27]

Answer:

P(0) = 0.6825

P(1) = 0.2607

Step-by-step explanation:

From the given information, the number of homicide is 107 and the total number of homicides per city –year is 280.

Let us denote the number of homicides per city-year as X.

The mean value, X is calculated as:

\begin{array}{c}\\{\rm{Mean}} = \frac{{107}}{{280}}\\\\ = 0.382\\\end{array}

Mean= 107/280 = 0.382

The mean number of homicides per city- year \left( {\lambda = \mu } \right)(λ=μ) is 0.382.

a. The probability that zero homicides is obtained is as below:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ -0.382}}{{\left( {0.382} \right)}^0}}}{{0!}}\\\\ = \frac{{\left( {0.6825 \right)\left( {\rm{1}} \right)}}{1}\\\\ = 0.6825\\\end{array}

P(X=0) = e −0.382 (0.382)⁰/1

= (0.6825)(1) /1

P(X=0) = 0.6825

Thus, the probability that zero homicides P(0) is 0.6825.

b. The probability that one homicides is obtained is as below:

\begin{array}{c}\\P\left( {X = 0} \right) = \frac{{{e^{ -0.382}}{{\left( {0.382} \right)}^0}}}{{0!}}\\\\ = \frac{{\left( {0.6825 \right)\left( {\rm{1}} \right)}}{1}\\\\ = 0.6825\\\end{array}

P(X=1) = e −0.382 (0.382)¹/1

= (0.6825)(0.382)/1

P(X=1) = 0.2607

Thus, the probability that zero homicides P(0) is 0.2607.

5 0

2 years ago

What is the greatest possible percent error in calculating the volume of a box measured to the nearest inch with sides of 5 in.,

Mashutka [201]

The answer is 20 percent

7 0

2 years ago

Read 2 more answers