What are the terms in the expression 2x−4y+8 ?

The city bus station runs buses every 15 minutes. Find the probability that a passenger who arrives at a random time will not ha

Paul [167]

The answer to your problem is A 1/3, is the possibility for a passenger who arrives at noon.

3 0

3 years ago

For the diagram below, form an equation and then solve it to calculate the value of x

Talja [164]

You apply the sum of interior angles formula ie. (n-2)180. n=number of sides
since a pentagon has 5 sides it will be (5-2)180=540.
now add everything. x-5+x-6+2x-7+x+2x-2=540.
Solve for x: 7x-20=540
7x=540+20
7x=560
x=560/7
x=80

8 0

3 years ago

I need help w this soon

Shtirlitz [24]

Answer:

18

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

Convert the following equations from point-slope form to slope-intercept form.

mojhsa [17]

<h2>Answer:</h2>

We need to solve both for y.

#9:

$y - 4 = 3(x - 1)\\\\y - 4 = 3x - 3\\\\y = 3x + 1$

#10:

$y - 1 = -4[x - (-3)]\\\\y - 1 = -4(x + 3)\\\\y - 1 = -4x - 12\\\\y = -4x - 11$

7 0

3 years ago

In a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer

stira [4]

Answer:

We conclude that 80% of patients stop smoking when given sustained care.

Step-by-step explanation:

We are given that in a program designed to help patients stop smoking. 192 patients were given to sustained care, and 80.2% of them were no longer smoking after one month.

Let p = percentage of patients stop smoking when given sustained care.

So, Null Hypothesis, $H_0$ : p = 80% {means that 80% of patients stop smoking when given sustained care}

Alternate Hypothesis, $H_A$ : p $\neq$ 80% {means that different from 80% of patients stop smoking when given sustained care}

The test statistics that would be used here One-sample z test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion of patients who stop smoking when given sustained care = 80.2%

n = sample of patients = 192

So, the test statistics = $\frac{0.802-0.80}{\sqrt{\frac{0.802(1-0.802)}{192} } }$

= 0.07

The value of z test statistics is 0.07.

Also, P-value of the test statistics is given by the following formula;

P-value = P(Z > 0.07) = 1 - P(Z $\leq$ 0.07)

= 1 - 0.52790 = 0.4721

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that 80% of patients stop smoking when given sustained care.

8 0

3 years ago