4y = -23 + 39
4y = 16
y = 4
Sub y = 4 into equation 2
7x + 6(4) = -39
7x + 24 = -39
7x = -39-24
7x = -63
x = -9
x = -9
y = 4
I = PRT
P = 8000
T = 4
R = 12%...turn to decimal...0.12
now we sub
I = (8000)(0.12)(4)
I = 3840 <==
Background: log (a . b) = log a + log b
hence log(10000x) = log 10000 + log x but we know that log 10000 = 4
= 4 + log x
therefore the graph of log(10000x) is just the graph of log x transposed by 4
The number of tests that it would take for the probability of committing at least one type I error to be at least 0.7 is 118 .
In the question ,
it is given that ,
the probability of committing at least , type I error is = 0.7
we have to find the number of tests ,
let the number of test be n ,
the above mentioned situation can be written as
1 - P(no type I error is committed) ≥ P(at least type I error is committed)
which is written as ,
1 - (1 - 0.01)ⁿ ≥ 0.7
-(0.99)ⁿ ≥ 0.7 - 1
(0.99)ⁿ ≤ 0.3
On further simplification ,
we get ,
n ≈ 118 .
Therefore , the number of tests are 118 .
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Answer:
$73.60
$345
simple interest = amount deposited x time x interest rate
600 + (600 x 0.055 x 5) = $765
600 + (600 x 0.055 x 5) > $2000
$765 $2000
He would not have $2000 in 5 years
Step-by-step explanation:
Total cost of items purchased = $75 + (2 x $8.50) = 92
If there is a 20% discount, he would pay (100 - 20%) 80% of the total cost =
0.8 x $92 = $73.60
commission earned = percentage commission x amount of sales
10% x $3450
= 0.1 x 3450 = 345
Amount he would have in his account = amount deposited + simple interest
simple interest = amount deposited x time x interest rate
600 x 0.055 x 5 = $165
Amount in his account in 5 years = $165 + 600 = $765
He would have less than $2000 in his account. he would have $765
