Answer:
12 credits per semester
Step-by-step explanation:
she has 92 in the end but she started off with 8 so your first step is to subtract 8 from 92 which gives you 84 once you get 84 you divide that by 12 you get 7 which means she got 12 credits in each of her 7 semeters
The confidence interval for a <span>(1−α)%</span> confidence level is given by
<span>
(<span>θ0</span>−<span>Z<span>α/2 </span></span><span>σ/√n</span>, <span>θ0</span>+<span>Z<span>α/2 </span></span><span>σ/√n</span>)
</span><span>θ0</span> is the measured statistic, <span>Z<span>α/2</span></span> is the cutoff/critical value, and <span>σ/<span>√n</span></span> is the standard error. σ is the population standard deviation (if known) or can be estimated by a sample standard deviation. n is the sample size.
The cutoff value depends on the test you wish to use, and <span>θ0</span><span> depends on the statistic you wish to estimate.</span>
Answer:
70x³ - 73x² - 155x + 168
Step-by-step explanation:
Given
(5x - 7)(2x + 3)(7x - 8) ← expand the first pair of factors using FOIL
= (10x² + x - 21)(7x - 8)
Each term in the second factor is multiplied by each term in the first factor, that is
10x² (7x - 8) + x(7x - 8) - 21(7x - 8) ← distribute the 3 parenthesis
= 70x³ - 80x² + 7x² - 8x - 147x + 168 ← collect like terms
= 70x³ - 73x² - 155x + 168
Base case: if <em>n</em> = 1, then
1² - 1 = 0
which is even.
Induction hypothesis: assume the statement is true for <em>n</em> = <em>k</em>, namely that <em>k</em> ² - <em>k</em> is even. This means that <em>k</em> ² - <em>k</em> = 2<em>m</em> for some integer <em>m</em>.
Induction step: show that the assumption implies (<em>k</em> + 1)² - (<em>k</em> + 1) is also even. We have
(<em>k</em> + 1)² - (<em>k</em> + 1) = <em>k</em> ² + 2<em>k</em> + 1 - <em>k</em> - 1
… = (<em>k</em> ² - <em>k</em>) + 2<em>k</em>
… = 2<em>m</em> + 2<em>k</em>
… = 2 (<em>m</em> + <em>k</em>)
which is clearly even. QED
The perimeter = 20 and AC = 8
Now as it is not mentioned which sides are equal of the isosceles triangle ABC,
We have two possible situations.
1)
If AC is the base
In that case AB = BC
Now AC = 8, AB = x , BC = x
So x + x + 8 = 20
2x + 8 = 20
2x = 12
x = 6
AB = BC = 6
2)
IF AC is not the base,
Then
AC = BC or AC = AB
So BC = 8 or AB = 8
If AB = AC = 8
Then
BC + 8 + 8 = 20
BC = 4
So there are two possible lengths of BC
Either it is BC = 8 or BC = 6 or BC = 4
The figure is attached for your reference.
