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2 years ago

Which of the following polynomials is in standard form?

Mathematics

2 answers:

ASHA 777 [7]2 years ago

6 0

Answer:

z5+z15-z3-4

hope it's correct

bogdanovich [222]2 years ago

3 0

Z25-z16+z14-4z is in standard form.

If the numbers after the variable are powers this is correct. In the future write it like this so people could help you easier (6z^2 instead of 6z2)

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A swimmer can swim 50.2 meters in 1 minute. How far can he swim in half an hour? *

belka [17]

Answer:

1506meters

Step-by-step explanation:

1hour-60minutes

which means half an hour is 30minutes,so it's more like they are asking you to find how far the swimmer can swim in 30minutes

50.2-1min

x -30min

x=50.2×30

x=1506meters

6 0

2 years ago

Read 2 more answers

A certain type of thread is manufactured with a mean tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilogram

azamat

Answer:

(a) The variance decreases.

(b) The variance increases.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the sample mean is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The standard deviation of sample mean is inversely proportional to the sample size, n.

So, if n increases then the standard deviation will decrease and vice-versa.

(a)

The sample size is increased from 64 to 196.

As mentioned above, if the sample size is increased then the standard deviation will decrease.

So, on increasing the value of n from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for n = 64 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7$

The standard deviation of the sample mean for n = 196 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4$

The standard deviation of the sample mean decreased from 0.7 to 0.4 when n is increased from 64 to 196.

Hence, the variance also decreases.

(b)

If the sample size is decreased then the standard deviation will increase.

So, on decreasing the value of n from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for n = 784 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2$

The standard deviation of the sample mean for n = 49 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8$

The standard deviation of the sample mean increased from 0.2 to 0.8 when n is decreased from 784 to 49.

Hence, the variance also increases.

6 0

2 years ago

) The GPA of accounting students in a university is known to be normally distributed. A random sample of 20 accounting students

katen-ka-za [31]

Answer:

The 95% confidence interval for the mean GPA of all accounting students at this university is between 2.5851 and 3.2549

Step-by-step explanation:

We are in posessions of the sample's standard deviation. So we use the student's t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 20 - 1 = 19

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975([tex]t_{975}$ ). So we have T = 2.0930