How much more did female bank tellers earn than male bank tellers as a percentage of male bank teller median earnings in 1969 (to the nearest tenth of a percent)?__ %

Answers

Answer 1

Answer:

Answer:

The additional percent that a median female bank teller earn than a median male bank teller in 1969 was 8.5% (to the nearest tenth of a percent)

Step-by-step explanation:

1. Let's review the information provided to us to answer the question correctly:

Female bank tellers median earnings in 1969 = US$ 4,190

Male bank tellers median earnings in 1969 = US$ 3,860

2. How much more did female bank tellers earn than male bank tellers as a percentage of male bank teller median earnings in 1969 (to the nearest tenth of a percent)?

For answering the question we will use the following formula:

Additional percent that a median female bank teller earn than a median male bank teller in 1969 = Female bank tellers median earnings in 1969/Male bank tellers median earnings in 1969 - 1

Replacing with the real values, we have:

Additional percent that a median female bank teller earn than a median male bank teller in 1969 = 4,190/3,860 - 1

Additional percent that a median female bank teller earn than a median male bank teller in 1969 = 1.085 - 1

Additional percent that a median female bank teller earn than a median male bank teller in 1969 = 0.085 = 8.5%

The additional percent that a median female bank teller earn than a median male bank teller in 1969 was 8.5%

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A home security system is designed to have a 99% reliability rate. Suppose that nine homes equipped with this system experience an attempted burglary. Find the probabilities of these events:_________.A. At least one alarm is triggered.
B. More than seven alarms are triggered.
C. Eight or fewer alarms are triggered.

Answers

Answer:

a) $P(X \geq 1) = 1-P(X<1) = 1-P(X=0)$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^(9-0)=1x10^(-18)$

And replacing we got:

$P(X \geq 1) =1 -1x10^(-18) \approx 1$

b) $P(X=7)=(9C7)(0.99)^7 (1-0.99)^(9-7)=0.003355$

$P(X=8)=(9C8)(0.99)^8 (1-0.99)^(9-8)=0.083047$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^(9-9)=0.913517$

And adding we got:

$P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992$

c) $P(X \leq 8) =1 -P(X>8) = 1-P(X=9)$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^(9-9)=0.913517$

And replacing we got:

$P(X \leq 8)= 1-0.913517=0.086483$

Step-by-step explanation:

Let X the random variable of interest "numebr of times that an alarm is triggered", on this case we now that:

$X \sim Binom(n=9, p=0.99)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^(n-x)$

Where (nCx) means combinatory and it's given by this formula:

$nCx=(n!)/((n-x)! x!)$

Part a

We want to find this probability:

$P(X \geq 1) = 1-P(X<1) = 1-P(X=0)$

$P(X=0)=(9C0)(0.99)^0 (1-0.99)^(9-0)=1x10^(-18)$

And replacing we got:

$P(X \geq 1) =1 -1x10^(-18) \approx 1$

Part b

$P(X \geq 7)= P(X=7) +P(X=8)+ P(X=9)$

$P(X=7)=(9C7)(0.99)^7 (1-0.99)^(9-7)=0.003355$

$P(X=8)=(9C8)(0.99)^8 (1-0.99)^(9-8)=0.083047$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^(9-9)=0.913517$

And adding we got:

$P(X \geq 7) = 0.003355+0.083047+0.913517 =0.99992$

Part c

$P(X \leq 8) =1 -P(X>8) = 1-P(X=9)$

$P(X=9)=(9C9)(0.99)^9 (1-0.99)^(9-9)=0.913517$

And replacing we got:

$P(X \leq 8)= 1-0.913517=0.086483$

QuestionEnter the exponential function using t (for time) as the independent variable to model the situation. Thenfind the value of the function after the given amount of time.A new savings account is opened with $400 and gains 3.5% yearly for 5 years.The exponential function that models the situation is y = MyAfter 5 years, the savings account has $

Answers

Since, it is an exponential function, thus this is a compound interest problem;

Where; the function is given as;

$\begin{gathered} A(t)=P(1+r)^t \n \text{Where A(t)= amount in the savings account at a time t} \n P=ca\text{ pital invested} \n r=\text{rate } \n t=\text{ time} \end{gathered}$

Thus, the function required is;

$\begin{gathered} A(t)=400(1+(3.5)/(100))^t \n A(t)=400(1.035)^t \end{gathered}$ $\begin{gathered} y=400(1.035)^t \n \text{Where t is the time} \end{gathered}$

After 5 years,

$\begin{gathered} A(5)=400(1.035)^5 \n A(5)=400(1.1877) \n A(5)=475.07 \end{gathered}$

The amount in the savings account after five years is $475.07

PLEASE HELP! I WILL MARK BRAINLIEST!

If ZY = 2x + 3 and WX = x+4, find WX.

Your answer will be so appreciated.

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i need more info on this ? to answer.

Consider the following hypothesis test:H 0: = 17H a: 17A sample of 40 provided a sample mean of 14.12. The population standard deviation is 4.a. Compute the value of the test statistic (to 2 decimals). (If answer is negative, use minus "-" sign.)b. What is the p-value (to 4 decimals)?c. Using = .05, can it be concluded that the population mean is not equal to 17? SelectYesNoItem 3Answer the next three questions using the critical value approach.d. Using = .05, what are the critical values for the test statistic (to 2 decimals)? ±e. State the rejection rule: Reject H 0 if z is Selectgreater than or equal togreater thanless than or equal toless thanequal tonot equal toItem 5 the lower critical value and is Selectgreater than or equal togreater thanless than or equal toless thanequal tonot equal toItem 6 the upper critical value.f. Can it be concluded that the population mean is not equal to 17?

Answers

Answer:

We conclude that the population mean is not equal to 17.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 17

Sample mean, $\bar{x}$ = 14.12

Sample size, n = 40

Alpha, α = 0.05

Population standard deviation, σ = 4

First, we design the null and the alternate hypothesis

$H_(0): \mu = 17\nH_A: \mu \neq 17$

We use Two-tailed z test to perform this hypothesis.

a) Formula:

$z_(stat) = \displaystyle\frac{\bar{x} - \mu}{(\sigma)/(√(n)) }$

Putting all the values, we have

$z_(stat) = \displaystyle(14.12 - 17)/((4)/(√(40)) ) = -4.5536$

b) P-value can be calculated from the standard z-table.

P-value = 0.0000

c) Since the p-value is less than the significance level, we reject the null hypothesis and accept the alternate hypothesis. Thus, the population mean is not equal to 17

d) Now, $z_(critical) \text{ at 0.05 level of significance } = \pm 1.96$

e) Rejection Rule:

We reject the null hypothesis if it is less than lower critical value and greater than the upper critical value

If the z-statistic lies outside the acceptance region which is from -1.96 to +1.96, we reject the null hypothesis.

f) Since the calculated z-stat lies outside the acceptance region, we reject the null hypothesis and accept the alternate hypothesis. Thus, the population mean is not equal to 17.

Final answer:

The test statistic is -1.78 and the p-value is 0.0761, indicating that we fail to reject the null hypothesis. Therefore, it cannot be concluded that the population mean is not equal to 17.

Explanation:

The test statistic can be calculated using the formula:

test statistic = (sample mean - population mean) / (population standard deviation / sqrt(sample size))

Plugging in the given values, we get:

test statistic = (14.12 - 17) / (4 / sqrt(40))

Calculating this gives us a test statistic value of -1.78.

The p-value can be calculated using the test statistic. We need to find the probability that a test statistic at least as extreme as -1.78 would occur assuming the null hypothesis is true. Using a standard normal distribution table or software, we find the p-value to be approximately 0.0761.

Since the p-value is greater than the significance level (alpha = 0.05), we fail to reject the null hypothesis. Therefore, we can conclude that there is not enough evidence to suggest that the population mean is not equal to 17.

Learn more about Hypothesis testing here:

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Help please! I’ll give brainliest!

Answers

Answer:the last one

Step-by-step explanation:

Which values from the specified set make up the solution set of the inequality?4n<16 ; {1,2,3,4}

Select ALL OF THE correct answers.

A. 1

B. 2

C. 3

D. 4