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nlexa [21]
3 years ago
6

What is the answer to janet makes homemade dolls. Currently, she produces 23 dolls per month. If she increase her production by

18% how many dolls would Janet produce each month?
Mathematics
1 answer:
34kurt3 years ago
4 0
27

This is because 18* of 23 is 4.14 which rounds to 4, and 23 + 4 = 27.
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If two polynomial equations have real solutions, then will the equation that is the result of adding, subtracting, or multiplyin
kirill115 [55]

No. A polynomial equation in one variablel ooks like P(x) = Q(x), where P and Q are polynomials.

Consider polynomial equations x^2 = 3 and x^2 = 1.

Obviously they have real solutions.

Subtract the two polynomial equations:

(x^2 - x^2) = (3 - 1)

0 = 2...

We get the polynomial equation 0 = 2. We call this a polynomial equation because single constants are also by definition polynomials.

Obviously 0 = 2 has no real solution.

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3 years ago
Explain what is meant by domain and range.
prohojiy [21]
The domain are the required input for a given function, while the range are the output or result from the function.
5 0
3 years ago
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5 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
3 years ago
Find the area of the triangle.<br> 10.<br> 15 cm<br> 16 cm
aev [14]
120
A= bh 1/2
15 x 16 = 240
240/2
120
4 0
3 years ago
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