<h3>Answer to Question 1:</h3>
AB= 24cm
BC = 7cm
<B = 90°
AC = ?
<h3>Using Pythagoras theorem :-</h3>
AC^2 = AB^2 + BC ^ 2
AC^2 = 24^2 + 7^2
AC^2 = 576 + 49
AC^2 = √625
AC = 25
<h3>Answer to Question 2 :-</h3>
sin A = 3/4
CosA = ?
TanA = ?
<h3>SinA = Opp. side/Hypotenuse</h3><h3> = 3/4</h3>
(Construct a triangle right angled at B with one side BC of 3cm and hypotenuse AC of 4cm.)
<h3>Using Pythagoras theorem :-</h3>
AC^2 = AB^2 + BC ^ 2
4² = AB² + 3²
16 = AB + 9
AB = √7cm
<h3>CosA = Adjacent side/Hypotenuse</h3>
= AB/AC
= √7/4
<h3>TanA= Opp. side/Adjacent side</h3>
=BC/AB
= 3/√7
Answer:
cosR = 
Step-by-step explanation:
assuming the right triangle has ∠ Q = 90° with legs 5 and 12
then this is a 5- 12- 13 right triangle with hypotenuse PR = 13 cm
then
cosR = 
= 
=
171/3 is 57. the three consecutive numbers are 56, 57 and 58.
244/4 is 61. the even integers are 58, 60, 62 and 64
Answer:
Step-by-step explanation:
Hello!
X: the lifespan of a new computer monitor of Glotech.
The average life is μ= 85 months and the variance δ²= 64
And a sample of 122 monitors was taken.
You need to calculate the probability that the sample mean is greater than 86.6 months.
Assuming that the variable has a normal distribution X~N(μ;δ²), then the distribution of the sample mean is X[bar]~N(μ;δ²/n)
To calculate this probability you have to work using the sampling distribution and the following formula Z= (X[bar]-μ)/δ/√n ~N(0;1)
P(X[bar]>86.6)= 1 - P(X[bar]≤86.6)
1 - P(Z≤(86.6-85)/(8/√122))= 1 - P(Z≤2.21)= 1 - 0.98645= 0.013355
The probability of the sample mean is greater than 0.013355
I hope this helps!
3/5*10 is the same as 3/5*10/1 and it should be 30/5 which is the same as 6.
So, 3/5*10 is 6
