In the equation Y+2/y+1 + 2y^2-y+3/y^2-1 = y-2/y-1 the value of a is: a)2 b)1 c)4

Corinne has a job selling magazines. She earns $7.50 per hour plus 20% of the total amount of her sales. She also gets an allowa

maw [93]

PART A
Coefficient
Coefficient h is 7.50
Coefficient s is 0.20

Variable
h and s

Constant
40

PART B
7.50h + 0.20s + 40
= 7.50(25) + 0.20(300) + 40
= 187.5 + 60 + 40
= 287.5
She earns $287.5

PART C
Yes, the coefficient of h would change to 9, the rests are still the same. Because she's no longer receive 7.50 per hour, and start earning 9 per hour, so the coefficient should change.
The expression would be
9h + 0.20s + 40

4 0

2 years ago

Plz help I don’t know the answer very important question

Igoryamba

Answer

D.26

Step-by-step explanation:

4 0

2 years ago

PLEASE HELP ME THANK YOU IF U DO

Andrews [41]

Answer:

B (r= 1/6s)

Step-by-step explanation:

If you just work out each option, you'll see that "s" multiplied by 1/6 gets r.

5 0

2 years ago

Simplify:<br>(ay<br>8<br>6.<br>5x - 25 30 - 6x

maw [93]

Answer:

-x - 2530

Step-by-step explanation:

5x - 2530 - 6x

5x - 6x = -x

-x - 2530

6 0

3 years ago

Read 2 more answers

Clarinex is a drug used to treat asthma. In clinical tests of this drug, 1655 patients were treated with 5- mg doses of Clarinex

bagirrra123 [75]

Answer:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

$p_v =P(z>3.363)=0.00039$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

Step-by-step explanation:

Data given and notation

n=1655 represent the random sample taken

$\hat p=0.021$ estimated proportion of interest

$p_o=0.012$ is the value that we want to test

$\alpha=0.01$ represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

$p_v$ represent the p value (variable of interest)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that true proportions is higher than 0.012.:

Null hypothesis: $p \leq 0.012$

Alternative hypothesis: $p > 0.012$

When we conduct a proportion test we need to use the z statisitic, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

The One-Sample Proportion Test is used to assess whether a population proportion $\hat p$ is significantly different from a hypothesized value $p_o$ .

Calculate the statistic

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.021 -0.012}{\sqrt{\frac{0.012(1-0.012)}{1655}}}=3.363$

Statistical decision

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.

The significance level provided $\alpha=0.01$ . The next step would be calculate the p value for this test.

Since is a right tailed test the p value would be:

$p_v =P(z>3.363)=0.00039$

So the p value obtained was a very low value and using the significance level given $\alpha=0.01$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of interest is significantly higher than 0.012 (1.2%)

3 0

3 years ago