The question in Englih is
<span>From a cardboard sheet 35 cm long and 20 cm wide, Masha cut out four squares of 1 dm2 each. Find the area of the cardboard residue. Answer the question in dm2
</span>
Step 1
<span>convert cm to dm
</span>we know that
1 cm is---------------> 0.10 dm
then
35 cm--------------> 3.5 dm
20 cm--------------> 2 dm
Step 2
find the area of the cardboard
Area=3.5*2=7 dm²
Step 3
find the area of the cardboard residue
Area=7-4*1=3 dm²
the answer is 3 dm²
<span>the answer in Russian
</span>
Шаг 1
преобразовать cm в дм
мы знаем это
1 cm ---------------> 0.10 дм
тогда
35 cm--------------> 3.5 дм
20 cm--------------> 2 дм
Шаг 2
найти область картона
Площадь=3.5*2=7 дм²
Шаг 3
<span>найти область остатка картона
</span>Площадь=7-4*1=3 дм²
ответ 3 дм²
ответ 3 дм²
Answer:
easy peasy lemon squeasy
Step-by-step explanation:
1. base of triangle: it is an equilateral triangle, so any side of the triangle you take is the base, (usually the largest side of the triangle is the base)
so that would be => 10 inches,
2. a property of an equilateral triangle is that , the median of the triangle coincides with the height of the triangle,
therefore, the height bisects the base and makes a right triangle with another side, therefore by using the Pythagoras theorem, we find it to be 8.66(approx)
now,
area of the triangle would be (putting the values we just found in the formula given)
=> [(10)(8.66)]/ 2
=>43.3 square inches
Should be 1/4 to my knowledge.
Answer:
The probability that all the five flights are delayed is 0.2073.
Step-by-step explanation:
Let <em>X</em> = number of domestic flights delayed at JFK airport.
The probability of a domestic flight being delayed at the JFK airport is, P (X) = <em>p</em> = 0.27.
A random sample of <em>n</em> = 5 flights are selected at JFK airport.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> and <em>p</em>.
The probability mass function of <em>X</em> is:
Compute the probability that all the five flights are delayed as follows:
Thus, the probability that all the five flights are delayed is 0.2073.
