Find the Antilog of 547.840

Select the ordered pair that is a solution to the function. (A point on the line.)

SIZIF [17.4K]

Answer:

The answer is c) (1,0)

Step-by-step explanation:

7 0

2 years ago

Martina started with 228 boxes of cookies and sold the same number each day for 4 days. After 4 days of selling, she was left wi

liq [111]

Answer:

$m = -32$

Step-by-step explanation:

Let x = days and y = box of cookies

So, we have:

$(x_1,y_1) = (0,228)$ -- When she started

$(x_2,y_2) = (4,100)$ --- After 4 days

Required

Determine the slope

The slope (m) is calculated using:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

$m = \frac{100 - 228}{4 - 0}$

$m = \frac{-128}{4}$

$m = -32$

<em>The slope is -32</em>

6 0

3 years ago

In a survey of 1000 women age 22 to 35 who work full time, 540 indicated that they would be willing to give up some personal in

morpeh [17]

Answer:

$z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.530$

$p_v =P(Z>2.530)=0.0057$

So 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 women indicated that they would be willing to give up some personal in order to make more money is significantly higher than 0.5.

The 95% confidence interval would be given by (0.509;0.571)

Step-by-step explanation:

Data given and notation

n=1000 represent the random sample taken

X=540 represent the women indicated that they would be willing to give up some personal in order to make more money

$\hat p=\frac{540}{1000}=0.54$ estimated proportion of women indicated that they would be willing to give up some personal in order to make more money

$p_o=0.5$ 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)

Concepts and formulas to use

We need to conduct a hypothesis in order to test the claim that the majority of woman age 22 to 35 who work full-time would be willing to give up some personal time for more money.:

Null hypothesis: $p \leq 0.5$

Alternative hypothesis: $p > 0.5$

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 $p_o$ .

Calculate the statistic

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

$z=\frac{0.54 -0.5}{\sqrt{\frac{0.5(1-0.5)}{1000}}}=2.530$

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 assumed is $\alpha=0.05$ . The next step would be calculate the p value for this test.

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

$p_v =P(Z>2.530)=0.0057$

So 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 women indicated that they would be willing to give up some personal in order to make more money is significantly higher than 0.5.

Confidence interval

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.54 - 1.96\sqrt{\frac{0.54(1-0.54)}{1000}}=0.509$

$0.54 + 1.96\sqrt{\frac{0.54(1-0.54)}{1000}}=0.571$

The 95% confidence interval would be given by (0.509;0.571)

5 0

3 years ago

At the beginning of the month, Kim had $66.78. Since then, she received three payments of $33.50 from her babysitting job. Kim's

cluponka [151]

Answer:

If Kim did not spend any money, she has $147.78 now.

Step-by-step explanation:

If Kim did not spend any money, the money she has now would be equal to the amount she had at the beginning of the month plus the three payments she received minus the amount she paid her sisters, which is:

(66.78+(3*33.50))-(2*9.75)=(66.78+100.5)-19.5=167.28-19.5=147.78

According to this, the answer is that if Kim did not spend any money, she has $147.78 now.

3 0

2 years ago

morgan has $28.75 on a gift card. she buys seven cups of coffee and has $20 left write and solve an equation to determine the co

Rainbow [258]

Step-by-step explanation:

28.75-20=8.75

8.75/7= 1.25

each cup costs $1.25

3 0

3 years ago

Find the Antilog of 547.840​

Find the Antilog of 547.840