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oee [108]
3 years ago
11

Sgagwhjdbwjjahavajansmkshsgsbsnkdjfhdbdbdjdqhahr

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
5 0

Answer:

huh?

Step-by-step explanation:

huh what does "sgagwhjdbwjjahavajansmkshsgsbsnkdjfhdbdbdjdqhahr" mean?

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Can someone help me .
insens350 [35]
Y-y1=m(x-x1)
m= slope =-3
y-y1=-3(x-x1),
y1=-7, x1=5
y--7=-3(x-5)
y+7=-3(x-5) this is C
7 0
3 years ago
Historically, the average time to service a customer complaint has been 3 days and the standard deviation has been 0.50 day. Man
kicyunya [14]
Oof I didn’t learn about this you can ask a class and go over it
6 0
2 years ago
The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to d
Hoochie [10]

Answer:

Step-by-step explanation:

Hello!

X: number of gypsy moths in a randomly selected trap.

This variable is strongly right-skewed. with a standard deviation of 1.4 moths/trap.

The mean number is 1.2 moths/trap, but several have more.

a.

The population is the number of moths found in traps places by the agriculture departments.

The population mean μ= 1.2 moths per trap

The population standard deviation δ= 1.4 moths per trap

b.

There was a random sample of 60 traps,

The sample mean obtained is X[bar]= 1

And the sample standard deviation is S= 2.4

c.

As the text says, this variable is strongly right-skewed, if it is so, then you would expect that the data obtained from the population will also be right-skewed.

d. and e.

Because you have a sample size of 60, you can apply the Central Limit Theorem and approximate the distribution of the sampling mean to normal:

X[bar]≈N(μ;σ²/n)

The mean of the distribution is μ= 1.2

And the standard deviation is σ/√n= 1.4/50= 0.028

f. and g.

Normally the distribution of the sample mean has the same shape of the distribution of the original study variable. If the sample size is large enough, as a rule, a sample of size greater than or equal to 30 is considered sufficient you can apply the theorem and approximate the distribution of the sample mean to normal.

You have a sample size of n=10 so it is most likely that the sample mean will have a right-skewed distribution as the study variable. The sample size is too small to use the Central Limit Theorem, that is why you cannot use the Z table to calculate the asked probability.

I hope it helps!

6 0
3 years ago
-5x +13y =-7<br><br> 5x +4y=24 <br><br> x=<br> y=
Dennis_Churaev [7]
X=4
Y=1
You already have the equation done from that equation you cross out the -5x and 5x then add 13+4 it gives you 17 the add -7+34 it gives you 17 that’s when you get x. Now ones you have x plug in the 1 into any equation I prefer the ones that don’t have negative numbers, ok so the plug in 1 into the Y and then it would be 5x+4(1)=24, so you know 4(1) is 4 so then subtract 24-4 which gives you 20 and then bring down 5x and then divide 24/5 and then you get what X is and it is 5.
3 0
3 years ago
In a​ poll, 37​% of the people polled answered yes to the question​ "Are you in favor of the death penalty for a person convicte
Kay [80]

Answer:

The number of people ​surveyed was 330.

Step-by-step explanation:

The (1 - <em>α</em>)% confidence interval for the population proportion is:

CI=\hat p\pm z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}

The margin of error for this interval is:

MOE= z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}

The information provided is:

\hat p=0.37\\MOE=0.05\\\text{Confidence level}=0.94\\\Rightarrow \alpha=0.06

The critical value of <em>z</em> for 94​% confidence level is, <em>z</em> = 1.88.

*Use a <em>z</em>-table.

Compute the value of <em>n</em> as follows:

MOE= z_{\alpha/2}\cdot \sqrt{\frac{\hat p(1-\hat p)}{n}}

      n=[\frac{z_{\alpha/2}\times \sqrt{\hat p(1-\hat p)}}{MOE}]^{2}

         =[\frac{1.88\times \sqrt{0.37(1-0.37)}}{0.05}]^{2}\\\\=(18.153442)^{2}\\\\=329.5475\\\\\approx 330

Thus, the number of people ​surveyed was 330.

5 0
2 years ago
Read 2 more answers
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