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snow_tiger [21]
3 years ago
8

On a factory floor, 30 out of every 140 toy robots is defective. What percent of you robots are defective? Round to the nearest

hundredth.
Mathematics
2 answers:
Gekata [30.6K]3 years ago
7 0
30 out of 140
that means the equation is 30/140
that equals .2142 . . .
that moves over the 21%
geniusboy [140]3 years ago
5 0

Answer: 21.43 %


Step-by-step explanation:

Given: On a factory floor, 30 out of every 140 toy robots is defective.

Thus, the total number of robots = 140

The number of defective robots = 30

Now the percent of robots which are defective= \frac{\text{defective robots}}{\text{total robots}}\times100

=\frac{30}{140}\times100

=21.42857\%\approx21.43\%

Hence, the percent  of robots which are defective = 21.43 %

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simple

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[BRAINLIEST] a. Nick deposits money in a Certificate of Deposit account (CD). The balance (in dollars) in his account t years af
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(1.06)0 = 1  and positive powers of 1.06 are larger than 1, thus the minimum value N(t) attains, if t≥0, is 400.

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A study was conducted and two types of engines, A and B, were compared. Fifty experiments were performed using engine A and 75 u
USPshnik [31]

Answer:

a) -6 mpg.

b) 2.77 mpg

c) The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

Step-by-step explanation:

To solve this question, we need to understand the central limit theorem, and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

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Gas mileage A: Mean 36, standard deviation 6, sample of 50:

So

\mu_A = 36, s_A = \frac{6}{\sqrt{50}} = 0.8485

Gas mileage B: Mean 42, standard deviation 8, sample of 50:

So

\mu_B = 42, s_B = \frac{8}{\sqrt{50}} = 1.1314

Distribution of the difference:

Mean:

\mu = \mu_A - \mu_B = 36 - 42 = -6

Standard error:

s = \sqrt{s_A^2+s_B^2} = \sqrt{0.8485^2+1.1314^2} = 1.4142

A. Find the point estimate.

This is the difference of means, that is, -6 mpg.

B. Find the margin of error

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1 - 0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1 - \alpha.

That is z with a pvalue of 1 - 0.025 = 0.975, so Z = 1.96.

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C. Construct the 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results(5 pts)

The lower end of the interval is the sample mean subtracted by M. So it is -6 - 2.77 = -8.77 mpg

The upper end of the interval is the sample mean added to M. So it is -6 + 2.77 = -3.23 mpg

The 95% confidence interval for the difference of population mean gas mileages for engines A and B and interpret the results, in mpg, is (-8.77, -3.23).

8 0
2 years ago
- 3/5 + - 3/4 can somebody help plz
Gnesinka [82]
<span>- 3/5 + - 3/4
= (-12/20) = (-15/20) 
= - 27/20
= - 1 7/20</span>
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arlik [135]

Answer:

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