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CaHeK987 [17]
3 years ago
8

Pls help like plsss and thx

Mathematics
2 answers:
Mandarinka [93]3 years ago
8 0
1.5 glad I could help!
geniusboy [140]3 years ago
3 0
The answer Would be 1.5
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Can someone plz help me with this hint it’s not D plz someone help me I’m not sure I understand this
Artyom0805 [142]

Answer:

The answer is C, or 3/5.

8 0
4 years ago
Solve for y:<br> y+5x=6;x= -1,0,1
AlekseyPX
When x = -1

y + 5x = 6

y = 6 - 5x

y = 6 - (5)(-1)

y = 6 - (-5)

y = 11

when x = 0

y = 6 - 5x

y = 6 - (5)(0)

y = 6 - 0

y = 6

when x = 1

y = 6 - 5x

y = 6 - (5)(1)

y = 6 - 5

y = 1
7 0
3 years ago
Read 2 more answers
Element X is a radioactive isotope such that every 25 years, its mass decreases by half. Given that the initial mass of a sample
Julli [10]
50*(.5)^(23/25)=26.43
7 0
3 years ago
According to a study conducted in one​ city, 25​% of adults in the city have credit card debts of more than​ $2000. A simple ran
Anvisha [2.4K]

Answer:

The sampling distribution of \hat p is <em>N</em> (0.25, 0.0354²).

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}=p

The standard deviation of this sampling distribution of sample proportion is:

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

Let <em>p</em> = proportion of adults in the city having credit card debts of more than​ $2000.

It is provided that the proportion of adults in the city having credit card debts of more than​ $2000 is, <em>p</em> = 0.25.

A random sample of size <em>n</em> = 150 is selected from the city.

Since <em>n</em> = 150 > 30 the Central limit theorem can be used to approximate the distribution of <em>p</em> by the Normal distribution.

The mean is:

\mu_{\hat p}=p=0.25

The standard deviation is:

\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.25(1-0.25)}{150}}=0.0354

Thus, the sampling distribution of \hat p is <em>N</em> (0.25, 0.0354²).

3 0
3 years ago
Random groups of 30 teachers were asked the starting annual salary for theirfirst teaching job. The sampling variability was low
a_sh-v [17]

Answer:

C. $69,000

Step-by-step explanation:

just did it on the test

8 0
3 years ago
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