Answer:
Note: The full question is attached as picture below
a) Hо : p = 0.71
Ha : p ≠ 0.71
<em>p </em>= x / n
<em>p </em>= 91/110
<em>p </em>= 0.83.
1 - Pо = 1 - 0.71 = 0.29.
b) Test statistic = z
= <em>p </em>- Pо / [√Pо * (1 - Pо ) / n]
= 0.83 - 0.71 / [√(0.71 * 0.29) / 110]
= 0.12 / 0.043265
= 2.77360453
Test statistic = 2.77
c) P-value
P(z > 2.77) = 2 * [1 - P(z < 2.77)] = 2 * 0.0028
P-value = 0.0056
∝ = 0.01
P-value < ∝
Reject the null hypothesis. There is sufficient evidence to support the researchers claim at the 1% significance level.
First, we must let:
x = number of tickets intended for adults
y = number of tickets intended for children.
a. Write in terms of x the number of tickets for children
Solution:
x + y = 28 ⇔ y = 28 - x (equation 1)
To answer in terms of x:
no. of tickets for tickets for children = 28 - x
b. the amount spent on tickets for adults
Solution: $30 is the cost of ticket per adult and there are x number of tickets intended for adults.
Therefore,
amount spent on ticket for adults = 30x
c. the amount spent on the tickets.
Solution:
$ 15 = cost of ticket per child
$ 30 = cost of ticket per adult
total amount spent on tickets = 30x + 15y ⇒ (equation2)
substitute equation 1 to equation 2.
(equation 1) y = 28 - x
(equation 2) total amount spent on tickets = 30x + 15y
total amount spent on tickets = 30x + 15(28-x)
total amount spent on tickets = 30x + 420 - 15x
total amount spent on tickets = 15x + 420
Hello there.
Which answer is the most reasonable estimation? Lance is organizing his baseball card collection into albums. Each album can hold up to 50 cards. He has 202 cards in his collection. About how many albums will Lance need? A. about 5 B. about 10 C. about 50 D. about 150
Answer: A) 5
The 4 in 42,672 is 40,000
And the 4 in 37,426 is 400
The answer is 100
