Y varies directly with x, and y is 16 when x is 32. Which equation represents this

Researchers fed mice a specific amount of Toxaphene, a poisonous pesticide, and studied their nervous systems to find out why To

Mkey [24]

Answer:

a. The mean refractory period= 1.85 and the standard error = 0.06455

b. 90% confidence interval for the mean absolute refractory period for all mice when subjected to the same treatment = 1.6981, 2.0019

c. Yes, the data give good evidence to support this theory

Step-by-step explanation:

a. The table below shows the calculations:

X (X-mean)^2

1.7 0.0225

1.8 0.0025

1.9 0.0025

2.0 0.0225

Total 7.4 0.05

Sample size: n=4

The mean is: \bar{x} = \frac{7.4}{4} = 1.85

The sample standard deviation, s = \sqrt{\frac{\sum \left ( x-\bar{x} \right )^{2}}{n-1}}=0.1291

The standard error, se= \frac{s}{\sqrt{n}}=\frac{0.1291}{2} = 0.06455

b. Degree of freedom: df = n-1 = 3

Critical value of t for 90% confidence interval is: 2.3534

The confidence interval is \bar{x}\pm t_{c}se = 1.85\pm 2.3534\cdot 0.06455=1.85\pm 0.1519 = (1.6981, 2.0019)

c. The Hypotheses are:

H_{0}:\mu=1.3,H_{1}:\mu>1.3

So the test statistics will be

t=\frac{\bar{x}-\mu}{s/\sqrt{n}}=8.52

The p-value is: 0.0017

We reject the null hypothesis because p-value is less than 0.05 . This indicates that the data gave good evidence to support this theory.

4 0

3 years ago

A ferry will safely accommodate 82 tons of passenger cars. Assume that the meanweight of a passenger car is 2 tons with standard

Shalnov [3]

Answer:

The probability that the maximum safe-weight will be exceeded is <u>0.0455 or 4.55%</u>.

Step-by-step explanation:

Given:

Maximum safe-weight of 37 cars = 82 tons

∴ Maximum safe-weight of 1 car (x) = 82 ÷ 37 = 2.22 tons (Unitary method)

Mean weight of 1 car (μ) = 2 tons

Standard deviation of 37 cars = 0.8 tons

So, standard deviation of 1 car is given as:

$\sigma=\frac{0.8}{\sqrt{37}}=0.13$

Probability that maximum safe-weight is exceeded, P(x > 2.22) = ?

The sample is normally distributed (Assume)

Now, let us determine the z-score of the mean weight.

The z-score is given as:

$z=\frac{x-\mu}{\sigma}\\\\z=\frac{2.22-2}{0.13}\\\\z=\frac{0.22}{0.13}=1.69$

Now, finding P(x > 2.22) is same as finding P(z > 1.69).

From the z-score table of normal distribution curve, the value of area under the curve for z < 1.69 is 0.9545.

But we need the area under the curve for z > 1.69.

So, we subtract from the total area. Total area is 1 or 100%.

So, $P(z > 1.69) = 1 - P(z < 1.69)$

$P(z>1.69)=1-0.9545=0.0455\ or\ 4.55\%$

Therefore, the probability that the maximum safe-weight will be exceeded is 0.0455 or 4.55%.

8 0

3 years ago

Which expressions are equivalent to the given expression?

expeople1 [14]

Answer:

It is the 2nd answer choice 4 check 5

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

Write 9x - 8 as a verbal expression <br><br> Please the fast as possible!

Natalka [10]

Answer:

8 subtracted from the product of 9 and x

Hope this helps!

4 0

2 years ago

You estimate that a baby pig weighs 21 lbs but it weighs 28 lbs what is the percent of error

lbvjy [14]

Your estimate of the baby pig's weight is 2 pounds. However the actual weight of it is only 16 pounds.
Let's solve for the percentage error of this.
=> 20 - 16 = 4 pounds
=> 4 / 16 = 0.25
=> 0.25 * 100% = 25%
Thus, you added 25% of the baby pig's weight

3 0

3 years ago

Read 2 more answers