What number is halfway between 100,000 and 200,000

If channels under a membrane of a neuron were blocked, what would be the consequence?

Scilla [17]

The membrane fluidity would be impaired and the neuron would undergo apoptosis . and the osmosis would be disrupted and the neuron would be unable to communicate. Also many chemicals would not be able to move into and out of the neuron, disrupting communication.d. the neuron would be deprived of oxygen and would undergo necrotic death.

8 0

4 years ago

4.

Zigmanuir [339]

Answer: The answer would be letter A

slope (m)=-9/2

intercept on x-axis=56/9

intercept on y-axis=28

Explanation:

18x+4y=112

(18x)/112+(4y)/112=1

(x/1)/(56/9)+y/28=1

intercept on x-axis=56/9

intercept on y-axis=28

now, 4y=-18x+112

y=-9/2x+28

so, slope (m)=-9/2

Step-by-step explanation:

Move all terms that don't contain y to the right side and solve.

y = 28 − (9 x) /(2)

7 0

3 years ago

Write an algebraic expression for<br><br> "10 less than h"<br><br> Answer (no spaces in answer)

Agata [3.3K]

Answer:

h-10

Step-by-step explanation:

just subtract 10 from h

7 0

3 years ago

Read 2 more answers

Question 14 (5 points)

Jobisdone [24]

Parallel = same slope
Slope = 1/2
Y = 1/2x + b
Plug in the point
0 = -1/2 + b, b = 1/2
Solution : y = 1/2x + 1/2

The answer is A

3 0

4 years ago

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

3 years ago