92-59=<br> hurry i need the answer

A new test is developed to test for a certain allergy, giving "positive" or "negative" results to indicate that a person does or

notsponge [240]

Answer:

the first blank is 82, the second blank is 8,910. I got that third blank wrong and it should be 7.1 I think. and the last block is should not. to find the third box all you need to do is:
82+990=1,072 then divide 82 by 1,072 and round to the nearest tenth

5 0

2 years ago

Using the given points and line, determine the slope of the line. (1, 2) and (2, 1) slope = 2 slope = 1 slope = -1

Bas_tet [7]

Answer:

-1

Step-by-step explanation:

Slope formula: (y₂ - y₁) / (x₂ - x₁)

= (2 - 1) / (1 - 2)

= -1

8 0

2 years ago

Thomas sent out 1500 surveys about the towns favorite pizza parlor. He had 245 returned in week 1, 173 in week 2, 185 in week 3,

musickatia [10]

1500 total surveys sent
Returned: 245 + 173 + 185 + 92 = 695

1500-695 = 805

7 0

3 years ago

Read 2 more answers

If the height of the two cones are in the ratio 1:3 and the radii of their base are in the ratio 4:1,then what is the ratio of t

BigorU [14]

2:3 is the ratio
....

3 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

92-59= hurry i need the answer