Doug entered a conoe race he rowed 3.5 miles in .5 hrs what is his average speed

Can someone write down the answers pls!!!!!!!!!!

snow_lady [41]

Step-by-step explanation:

6b x>=1

6c y>7

7a x>3

7b -2<=2x<6

-1<=x<3

answer:-1,0,1,2,3

7 0

3 years ago

The population of a small town is 39,668 and has been continuously declining at a 1.5% a year what will the population be in 13

anyanavicka [17]

Answer: 31, 932.74

Step-by-step explanation:

The population of the small town = 39,668

Yearly rate of decline = 1.5%

Therefore, to calculate 1.5% of 39,668 = 1.5 x 39,668

------------------

100

= 595.02

So, the population is declining by 595.02 on yearly basis.

To ascertain what will the population would have declined to in the next 13 years,

We multiply , = 595.02 x 13

=7,735.26

Now population in 13 years = 39,668 - 7,735.26

= 31,932.74

6 0

4 years ago

A local company makes a candy that is supposed to weigh 1.00 ounces. A random sample of 25 pieces of candy produces a mean of 0.

AysviL [449]

Answer:

$n=(\frac{2.58(0.004)}{0.001})^2 =106.50 \approx 107$

So the answer for this case would be n=107 rounded up to the nearest integer

Step-by-step explanation:

Information given

$\bar X= 0.996$ the sample mean

$s=0.004$ the sample deviation

$n =25$ the sample size

Solution to the problem

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =0.001 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

We can use as estimator of the real population deviation the sample deviation for this case $\hat \sigma = s$ / The critical value for 99% of confidence interval is given by $z_{\alpha/2}=2.58$ , replacing into formula (b) we got:

$n=(\frac{2.58(0.004)}{0.001})^2 =106.50 \approx 107$

So the answer for this case would be n=107 rounded up to the nearest integer

6 0

3 years ago

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 $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

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.

Now, find the margin of error M as such

$M = zs = 1.96*1.4142 = 2.77$

The margin of error is of 2.77 mpg

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

3 years ago

What is AE? Please help me

shepuryov [24]

The two triangles are similar, so will have a direct scale factor. We can find this by using the given measurements, 10÷8=1.25. So the scale factor from the smaller to the bigger triangle is 1.25. This means that 1.25×(x+6)=AE. We can solve this as we would with a regular linear equation:
1.25(x+6)=2x+6
1.25x+7.5=2x+6
1.5=0.75x
2=x
∴ AE is 2x+6=10

4 0

3 years ago