A school bought several pairs of skis, each costing $40 and several pairs of skates, each costing $50. the cost of the whole pur
1 answer:
You might be interested in
Answer
Given,
Mass of bag = 6 Kg
Area covered by the bag = 2.5 x 5 = 12.5 m²
a) Bags required to cover area = 7.5 x 10.2
= 76.5 m²
Number of bags, N =
N = 6.12 = 7 (approx.)
7 Bags of 6 Kg fertilizers will be needed to cover the area.
b) Cost of 6 Kg fertilizers = $15.50
Cost of 7 bags of 6 Kg fertilizers = 7 x $15.50
= $108.50.
Answer:
i) 60°
ii) 14.7 cm
Step-by-step explanation:
length of Arc BA = 8.4 cm
length of radius = 8 cm
angle AOC = 45°
BO = AO = CO = 8 cm
<u>i) What is the measure of Angle BOA </u>
= length of Arc BA / Radius
= 8.4 / 8 = 1.05 radians = 60.16° ≈ 60°
<u>ii) what is the length of arc BAC</u>
= radius * angle BAC *
angle BAC = angle AOC + angle BOA
= 45° + 60° = 105°
hence length of arc BAC
= 8 * 105 *
≈ 14.7 cm
Answer:
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) We can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval
(3) A survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.
Step-by-step explanation:
We are given that a simple random sample of 331 American adults who do not have a four-year college degree and are not currently enrolled in school, 48% said they decided not to go to college because they could not afford school.
Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;
P.Q. = ~ N(0,1)
where, = sample proportion of Americans who decide to not go to college = 48%
n = sample of American adults = 331
p = population proportion of Americans who decide to not go to
college because they cannot afford it
<em>Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.</em>
<em />
<u>So, 90% confidence interval for the population proportion, p is ;</u>
P(-1.645 < N(0,1) < 1.645) = 0.90 {As the critical value of z at 5% level
of significance are -1.645 & 1.645}
P(-1.645 < < 1.645) = 0.90
P( <
<
) = 0.90
P( < p <
) = 0.90
<u>90% confidence interval for p</u> = [ ,
]
= [ ,
]
= [0.4348, 0.5252]
(1) Therefore, a 90% confidence interval for the proportion of Americans who decide to not go to college because they cannot afford it is [0.4348, 0.5252].
(2) The interpretation of the above confidence interval is that we can be 90% confident that the proportion of Americans who choose not to go to college because they cannot afford it is contained within our confidence interval.
3) Now, it is given that we wanted the margin of error for the 90% confidence level to be about 1.5%.
So, the margin of error =
= 54.79
n =
n = 3001.88 ≈ 3002
Hence, a survey should include at least 3002 people if we wanted the margin of error for the 90% confidence level to be about 1.5%.