Answer:
about 616 in²
Step-by-step explanation:
The area of a sphere is given by ...
A = πd²
For a ball 14 inches in diameter, the area is ...
A = π(14 in)² = 196π in²
A ≈ 616 in²
The area of the required plastic is about 616 square inches.
9514 1404 393
Answer:
(i) x° = 70°, y° = 20°
(ii) ∠BAC ≈ 50.2°
(iii) 120
(iv) 300
Step-by-step explanation:
(i) Angle x° is congruent with the one marked 70°, as they are "alternate interior angles" with respect to the parallel north-south lines and transversal AB.
x = 70
The angle marked y° is the supplement to the one marked 160°.
y = 20
__
(ii) The triangle interior angle at B is x° +y° = 70° +20° = 90°, so triangle ABC is a right triangle. With respect to angle BAC, side BA is adjacent, and side BC is opposite. Then ...
tan(∠BAC) = BC/BA = 120/100 = 1.2
∠BAC = arctan(1.2) ≈ 50.2°
__
(iii) The bearing of C from A is the sum of the bearing of B from A and angle BAC.
bearing of C = 70° +50.2° = 120.2°
The three-digit bearing of C from A is 120.
__
(iv) The bearing of A from C is 180 added to the bearing of C from A:
120 +180 = 300
The three-digit bearing of A from C is 300.
Answer:
1
Step-by-step explanation:

Using BODMAS we'll have to solve what is in the bracket first

Multply -7 by 5 still using the rule of BODMAS

<h2>
<em>OladipoSeun</em><em>♡˖꒰ᵕ༚ᵕ⑅꒱</em></h2>
Answer:
E. Approximately normal with standard deviation less than 0.7 sibling
Step-by-step explanation:
To solve this question, we use the Central Limit theorem.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation 
In this problem, we have that:
Skewed right distribution, with 
Sampling distribution of the sample mean for samples of size 100
By the Central Limit Theorem, they will be approximately normal, with mean
, and standard deviation 
So the correct answer is:
E. Approximately normal with standard deviation less than 0.7 sibling