Find the equation of perpendicular bisector of the lines with the given word (1,2) and (5,4)

Jaxon has $65 in his savings account. He deposits $15 every week. His father also deposits $25 into the account every time Jaxon

Grace [21]

Part A:
A coefficient can be either '15' or '25'.
A variable can be either 'w' or 'm'.
A constant is 65.

Part B:
Simply substitute, or plug-in, the numbers and solve.
***Step 1:
65+15w+25m --> 65+15(20)+25(3)
You do this because you are substituting the 'w' for the number
of weeks that Jaxon saved up for, which is 20, and the 'm' for the number
of times that Jaxon mowed the lawn, which is 3.
***Step 2:
65+15(20)+25(3) --> 65+300+75
Begin to solve, using PEMDAS, or whichever acronym you learned.
Remember, if you are using PEMDAS, recall that the order is Parenthesis,
Exponents, Multiplication/Division (whichever comes first), and
Addition/Subtraction (whichever comes first). Here, I checked for parenthesis.
I did find parenthesis, however, they do not have any expressions inside of
them, meaning that these parenthesis are for multiplying, and not for stating
order. So, you skip parenthesis. Next, you check for exponents, which you
find none of, so you skip over that. Now, we get to multiplying/dividing, so
you multiply the 15 and the 20 to get 300, and the 25 and 3 to get 75.
***Step 3:
65+300+75 --> 440
Now, we get to addition. You simply add everything up to get your final
answer: $440.

Part C:
If Jaxon had $75, then yes, the coefficients would change.
By subtracting $65 from $75, we can see that the total amount of money
from Jaxon's deposits and his lawn-mowing money is $10. Jaxon already
deposits $15 a week, meaning that, while using the current equation, Jaxon
CANNOT have $75 in his bank account. We can change the equation
so that Jaxon is able to have $75 in his savings account. You can change
the coefficient of 15 to 10, and the other coefficient of 25 to 0.
Now Jaxon is able to have $75 in his savings account.

8 0

3 years ago

. In a study of air-bag effectiveness it was found that in 821 crashes of midsize cars equipped with air bags, 46 of the crashes

jok3333 [9.3K]

Answer:

P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584

0.00885... < 0.01

The test statistic of 46 is significant

There is sufficient evidence to reject H₀ and accept H₁

Air bags are more effective as protection than safety belts

Step-by-step explanation:

821 crashes

46 hospitalisations where car has air bags

7.8% or 0.078 probability of hospitalisations in cars with automatic safety belts

α = 0.01 or 1% ← level of significance

One-tailed test

We are testing whether hospitalisations in cars with air bags are less likely than in a car with automatic safety belts;

The likelihood of hospitalisation in a car with automatic safety belts, we are told, is 7.8% or 0.078;

So we are testing if hospitalisations in cars with air bags is less than 0.078;

So, firstly:

Let X be the continuous random variable, the number of hospitalisations from a car crash with equipped air bags

X~B(821, 0.078)

Null hypothesis (H₀): p = 0.078

Alternative hypothesis (H₁): p < 0.078

According to the information, we reject H₀ if:

P(X ≤ 46 | X~B(821, 0.078)) < 0.01

To find P(X ≤ 46) or equally P(X < 47), it could be quite long-winded to do manually for this particular scenario;

If you are interested, the manual process involves using the formula for every value of x up to and including 46, i.e. x = 0, x = 1, x = 2, etc. until x = 46, the formula is:

$P(X = r) = nCr * p^{r} * (1 - p)^{n - r}$

You can find binomial distribution calculators online, where you input n (i.e. the number of trials or 821 in this case), probability (i.e. 0.078) and the test statistic (i.e. 46), it does it all for you, which gives:

P(X ≤ 46 | X~B(821, 0.078)) = 0.00885745584

Now, we need to consider if the condition for rejecting H₀ is met and recognise that:

0.00885... < 0.01

There is sufficient evidence to reject H₀ and accept H₁.

To explain what this means:

The test statistic of 46 is significant according to the 1% significance level, meaning the likelihood that only 46 hospitalisations are seen in car crashes with air bags in the car as compared to the expected number in car crashes with automatic safety belts is very unlikely, less than 1%, to be simply down to chance;

In other words, there is 99%+ probability that the lower number of hospitalisations in car crashes with air bags is due to some reason, such as air bags being more effective as a protective implement than the safety belts in car crashes.

5 0

2 years ago

While watching a movie, Ned, Fred, and Jed shared a bowl of popcorn. Ned ate 1/2 of the popcorn, Fred ate 1/4 of the popcorn, an

Charra [1.4K]

1/8 bowl.
1/2 = 4/8
1/4 = 2/8
1/8 = 1/8

4/8+2/8+1/8 = 7/8
1 - 7/8 = 1/8

6 0

2 years ago

Read 2 more answers

Help find the distance between two points?

miv72 [106K]

Answer:

4