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Diano4ka-milaya [45]
2 years ago
6

The average grade of an English class rose fro 70 to 85, what is the approximate percent increase?

Mathematics
1 answer:
hodyreva [135]2 years ago
5 0

Answer:

Step-by-step explanation:

85 - 70 = 15

15/70 = 0.21428571428

Round to .21

Move decimal over twice to the right

21%

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7 (8.02) need answer asap
timama [110]
Both groups are similar.
7 0
3 years ago
Can someone answer this correctly pls
AleksandrR [38]

Answer:

i = 45% and ii = 47% (Rounded)

Step-by-step explanation:

Simply divide 27 by 60 to get i:

27/60=.45

Convert to a percent chance.

.45=45%

Then divide 395 by 840 to get ii:

395/840=.47

Convert to a percent chance.

.47=47%

8 0
2 years ago
Helppppp PLEASEEEEE VERY ARGENTTTT​
vekshin1

Answer:

D; x -y = 22

3x + 2y = 246

Step-by-step explanation:

Let us have the bigger number as x and the smaller as y

The difference between the two is 22

Thus, we have it that;

x - y = 22

twice the smaller number; 2(y) added to thrice the larger 3(x) equals 246

3x + 2y = 246

So we have the two equations as;

x -y = 22

3x + 2y = 246

4 0
3 years ago
4.One attorney claims that more than 25% of all the lawyers in Boston advertise for their business. A sample of 200 lawyers in B
AleksAgata [21]

Answer:

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

p_v =P(Z>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

Step-by-step explanation:

1) Data given and notation  

n=200 represent the random sample taken

X=63 represent the lawyers had used some form of advertising for their business

\hat p=\frac{63}{200}=0.315 estimated proportion of lawyers had used some form of advertising for their business

p_o=0.25 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that more than 25% of all the lawyers in Boston advertise for their business:  

Null hypothesis:p\leq 0.25  

Alternative hypothesis:p > 0.25  

When we conduct a proportion test we need to use the z statistic, 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.

3) Calculate the statistic  

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

z=\frac{0.315 -0.25}{\sqrt{\frac{0.25(1-0.25)}{200}}}=2.123  

4) 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=0.05. 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>2.123)=0.0169  

The p value obtained was a very low value and using the significance level given \alpha=0.05 we have p_v so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the proportion of lawyers had used some form of advertising for their business is significantly higher than 0.25 or 25% .  

8 0
3 years ago
Round off the nearest place value 75,432,387
liberstina [14]
Since you'd didn't specify what place value I'm assuming it's the whole number so the answer is: 80,000,000
7 0
3 years ago
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