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Gelneren [198K]
3 years ago
15

I need help I got this wrong

Mathematics
1 answer:
wlad13 [49]3 years ago
3 0
<h2>Solution : </h2>

y = -2|x +1|

→ |x + 1| = -y/2

→ x+1 = y/2

and

x + 1 = -y/2

→ x + 1 belongs to R

→ x belongs to R

Domain = R

Range = y = -2(R)

→( -∞ , 0)

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Fill in the blank. A​ _______ variable is a variable that has a single numerical​ value, determined by​ chance, for each outcome
sineoko [7]

Answer:

Random variable

Step-by-step explanation:

The reason it is a random variable is because a, the definition fits, and b you can use context clues as well, such as 'determined by chance' which is another example of random! So, the answer is random, because random variables are determined by chance. Hope this helped.

4 0
3 years ago
Read 2 more answers
Factor 10c^2-60cd +80d^2
ozzi
10c^2-60cd +80d^2

(2c - 8d) (5c -10d)

this  \\ is \\ because...

2c*5c=10c^2

2c * -10d=-20cd \\ -8d*5c=-40cd \\ -20cd-40cd=-60cd

-8d * -10d = 80d
5 0
3 years ago
If 12 cows produce 70 gallons of milk, how many gallons of milk would 42 cows produce?
I am Lyosha [343]
This is a proportions question
  12        42
------  =  -------
  70          ?

Cross multiply

70x42=2940
12x?=2940
294/12=245

42 cows can produce 245 gallons of milk.
So the answer is B.

Hope I helped!

4 0
3 years ago
Read 2 more answers
Determine whether the vectors u and v are parallel, orthogonal, or neither.
Radda [10]
To check if two vectors are orthogonal(perpendicular), simply check their dot product, if their dot product is 0, then they're perpendicular, let's check.

\bf \ \textless \ 7,-4\ \textgreater \ \cdot \ \textless \ -28,16\ \textgreater \ \implies (7\cdot -28)+(-4\cdot 16)\implies -196-64&#10;\\\\\\&#10;\boxed{-264}\impliedby \textit{nope, no orthogonal}

to check if two vectors are parallel, simply check their slope by doing a b/a check, if the slopes are the same, then they're indeed parallel to each other, let's check.

\bf \textit{slope of \underline{u} }\cfrac{7}{-4}\implies \boxed{-\cfrac{7}{4}}\qquad \qquad \textit{slope of \underline{v} }\cfrac{-28}{16}\implies \boxed{-\cfrac{7}{4}}

well, there you have it, the slopes are the same.
4 0
4 years ago
Historically, the proportion of people who trade in their old car to a car dealer when purchasing a new car is 48%. Over the pre
choli [55]

Answer:

z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

Step-by-step explanation:

Data given and notation

n=115 represent the random sample taken

X=46 represent the number of people that have traded in their old car.

\hat p=\frac{46}{115}=0.4 estimated proportion of people that have traded in their old car

p_o=0.48 is the value that we want to test

\alpha=0.1 represent the significance level

Confidence=90% or 0.9

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the proportion is less than 0.48.:  

Null hypothesis:p\geq 0.48  

Alternative hypothesis:p < 0.48  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.4 -0.48}{\sqrt{\frac{0.48(1-0.48)}{115}}}=-1.717  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.1. The next step would be calculate the p value for this test.  

Since is a left tailed test the p value would be:  

p_v =P(z  

So the p value obtained was a very low value and using the significance level given \alpha=0.1 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of people that have traded in their old car is lower than 0.48 or 48%.  

4 0
4 years ago
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