All categories

3 years ago

Plzz help and explain ur answer i might give brainliest

Mathematics

1 answer:

adoni [48]3 years ago

3 0

Answer:

irrational

Step-by-step explanation:

π stands for the ratio of the circumference to the diameter of a circle, an irrational number near 3.14159. Multiplying any irrational number by a rational number (-11, for example) gives an irrational result.

-11π is irrational

You might be interested in

How do I solve this equation?

kow [346]

Answer: 2 * sqrt(2)

I attached an image to show you what I mean

Step-by-step explanation:

First add 20 on both sides

5x^(2 / 3) = 10

divide both sides by 5

x^(2 / 3) = 2

Now be reminded that the power of 2/3 is the same as x^2 cube root (to the third root)

So we now cube both sides to get rid of the cube root

x^2 = 2^3

which simplifies to x^2 = 8

We now just find the square root of both sides

so x = sqrt(8)

square root of 8 is simply sqrt(4 * 2) = sqrt(4) * sqrt(2) = 2 * sqrt(2)

So x = 2 * sqrt (2)

8 0

3 years ago

3-74. Mentally calculate the following products. Use the rule for decimal multiplication to write an equation in

dalvyx [7]

B. -.0006

C. .28

Multiply 3 and 2 and you would get 6 then move the decimal 4 places to the left and you would get .0006

Multiply 7 and 4 and you would get 28 then move the decimal places 2 spaces to the left and you would get .28

8 0

3 years ago

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

3 years ago

Assume that we want to construct a confidence interval. Do one of the following, as appropriate:_________.

andreyandreev [35.5K]

Answer:

B. z Subscript alpha divided by 2 zα/2 = 1.96.

Step-by-step explanation:

We are given that we want to construct a confidence interval. For this, the summary statistics for randomly selected weights of newborn girls:

n = 236, $\bar x$ = 30.3 hg, s = 7.2 hg. The confidence level is 95%.

As we can clearly see here that the population standard deviation is unknown and the sample size is also very large.

It has been stated that when the population standard deviation is unknown, we should use t-distribution but since the sample size is very large so we can use z distribution also as it is stated that at very large samples; the t-distribution corresponds to the z-distribution.

Here, $\alpha$ = level of significance = 1 - 0.95 = 0.05 or 5%