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Andreyy89
3 years ago
15

What's the slope of y=20+15x

Mathematics
2 answers:
Ivenika [448]3 years ago
5 0
The slope of this would be 15
Leviafan [203]3 years ago
5 0

Answer:

Slope = 15

Step-by-step explanation:

Comparing with y = mx + c

m = 15

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$4.95 chocolate bars are now on sale for 35% off. what will the chocolate bars cost now? (show work)
Tema [17]
= 4.95 X .65 = 3.2175 ( 3.22).
4.95 = 100% of the price
100 less 35 = 65
4.95 X .65% = 3.2175 or 3.22
6 0
3 years ago
40 POINTS!!!!!!!! The length of a rectangle is 7 millimeters longer than its width. The perimeter is more than 62 millimeters.
Arte-miy333 [17]

a) L = W + 7  

b) 2 ( L + W ) > 62  

c) 2 ( W + 7 + W ) > 62  

2w+14+2w>62  

4w+14>62  

4w > 48  

w > 12

8 0
3 years ago
In the past, the average age of employees of a large corporation has been 40 years. Recently, the company has been hiring older
Viktor [21]

Answer:

p_v =P(t_{(63)}>2.5)=0.0075  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.  

Step-by-step explanation:

1) Data given and notation  

\bar X=45 represent the mean height for the sample  

s=16 represent the sample standard deviation for the sample  

n=64 sample size  

\mu_o =40 represent the value that we want to test

\alpha=0.05 represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean age is higher than 40 years, the system of hypothesis would be:  

Null hypothesis:\mu \leq 40  

Alternative hypothesis:\mu > 40  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{45-40}{\frac{16}{\sqrt{64}}}=2.5    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=64-1=63  

Since is a one right tailed test the p value would be:  

p_v =P(t_{(63)}>2.5)=0.0075  

Conclusion  

If we compare the p value and the significance level given \alpha=0.05 we see that p_v so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can conclude that the mean age is significantly higher than 45 years at 5% of significance.  

6 0
2 years ago
How many square feet of outdoor carpet will we need for this hole
Leokris [45]
Can you describe it a little more- I have no clue how I’m suppose to solve that.
3 0
2 years ago
How many different linear arrangements are there of the letters a, b,c, d, e for which: (a a is last in line? (b a is before d?
inna [77]
A) Since a is last in line, we can disregard a, and concentrate on the remaining letters.
Let's start by drawing out a representation:

_ _ _ _ a

Since the other letters don't matter, then the number of ways simply becomes 4! = 24 ways

b) Since a is before d, we need to account for all of the possible cases.

Case 1: a d _ _ _ 
Case 2: a _ d _ _
Case 3: a _ _ d _
Case 4: a _ _ _ d

Let's start with case 1.
Since there are four different arrangements they can make, we also need to account for the remaining 4 letters.
\text{Case 1: } 4 \cdot 4!

Now, for case 2:
Let's group the three terms together. They can appear in: 3 spaces.
\text{Case 2: } 3 \cdot 4!

Case 3:
Exactly, the same process. Account for how many times this can happen, and multiply by 4!, since there are no specifics for the remaining letters.
\text{Case 3: } 2 \cdot 4!

\text{Case 4: } 1 \cdot 4!

\text{Total arrangements}: 4 \cdot 4! + 3 \cdot 4! + 2 \cdot 4! + 1 \cdot 4! = 240

c) Let's start by dealing with the restrictions.
By visually representing it, then we can see some obvious patterns.

a b c _ _

We know that this isn't the only arrangement that they can make.
From the previous question, we know that they can also sit in these positions:

_ a b c _
_ _ a b c

So, we have three possible arrangements. Now, we can say:
a c b _ _ or c a b _ _
and they are together.

In fact, they can swap in 3! ways. Thus, we need to account for these extra 3! and 2! (since the d and e can swap as well).

\text{Total arrangements: } 3 \cdot 3! \cdot 2! = 36
7 0
2 years ago
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