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3 years ago

Round this number to the nearest tenth. 52.85

Mathematics

1 answer:

k0ka [10]3 years ago

6 0

Answer:

52.9

numbers 0-4 the number stays the same

numbers 5-9 the number goes up one

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A coin was flipped 150 times. The results of the experiment are shown in the following table:

Basile [38]

<h2>Answer:</h2>

The statement that best describes the experimental probability of getting heads is:

It is 10% higher than the theoretical probability.

<h2>Step-by-step explanation:</h2>

We know that the theoretical probability of getting head is: 0.5

and the theoretical probability of getting tail is: 0.5

Also, the results of experiments are:

Heads Tails

90 60

This means that the experimental probability of getting head is: 90/150=3/5=0.6

and the experimental probability of getting tail is: 60/150=2/5=0.4

Hence, the experimental and theoretical probability of getting head is not same or equal.

Also, the experimental probability of getting head is more than the theoretical probability of getting head and the percent increase is calculated as:

$(0.6-0.5)\times 100\\\\\\=(0.1)\times 100\\\\=10\%$

8 0

3 years ago

Read 2 more answers

Write the slope intercept form of a line containing the y - intercept (0, -3) with a slope of –1.

Brums [2.3K]

Answer:

y = -x - 3

Step-by-step explanation:

4 0

3 years ago

9 1/5 into an improper fraction

vaieri [72.5K]

Answer:

46/5 hope it helps!

Step-by-step explanation:

9x5= 45

45+1= 46

46/5

7 0

3 years ago

(2pm^-1q^0)^-4 • 2m ^-1 p^3 / 2pq^2

Montano1993 [528]

Answer:

$\dfrac{m^3}{16p^2q^2}$

Step-by-step explanation:

Given:

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}$

$m^{-1}=\dfrac{1}{m}$

$q^0=1$

$2pm^{-1}q^0=2p\cdot \dfrac{1}{m}\cdot 1=\dfrac{2p}{m}$

$(2pm^{-1}q^0)^{-4}=\left(\dfrac{2p}{m}\right)^{-4}=\left(\dfrac{m}{2p}\right)^4=\dfrac{m^4}{(2p)^4}=\dfrac{m^4}{16p^4}$

$m^{-1}=\dfrac{1}{m}$

$2m^{-1} p^3=2\cdot \dfrac{1}{m}\cdot p^3=\dfrac{2p^3}{m}$

$\dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{\frac{2p^3}{m}}{2pq^2}=\dfrac{2p^3}{m}\cdot \dfrac{1}{2pq^2}=\dfrac{p^2}{mq^2}$

$(2pm^{-1}q^0)^{-4}\cdot \dfrac{ 2m^{-1} p^3}{2pq^2}=\dfrac{m^4}{16p^4}\cdot \dfrac{p^2}{mq^2}=\dfrac{m^3}{16p^2q^2}$

8 0

3 years ago

The world's population has grown at an average rate of 1.9 percent per year since 1945. There were approximately 4 billion peopl

kkurt [141]

Answer: $f(t) = 4000000000( 1 .019)^t$

Step-by-step explanation:

Here, According to the question, The population of the world is 4 billion (approx).

Again According to the question, The world's population has grown at an average rate of 1.9 percent per year since 1945.

Therefore, after 1975 the growth rate will remains same.

Thus, the population of the world after t years,

$f(t) = 4000000000( 1+\frac{1.9}{100} )^t$ (By the formula $A = P(1+\frac{r}{100})^t$ )

⇒ $f(t) = 4000000000( 1.019 )^t$

Where 1.019 is the growth rate and 4 billion is the initial population.

6 0

3 years ago