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son4ous [18]
2 years ago
14

Ashley is making $500 a week at her job. She receives a 20% pay raise. A year later she gets a 5% pay cut. How much does she mak

e per week now?
Mathematics
1 answer:
notsponge [240]2 years ago
8 0

Answer:

Assuming Ashley didn't​ take any week off. Ashley earning in the first year:

500*52=$26000

She earns the second year:

26000 * 1.2 = $31200

She spent 5%

31200 * 0.05 = $1560

The remaining yearly earning:

31200 - 1560 = $29640

Weekly earning:

29640 ÷ 52 = $570


Read more on Brainly.com - brainly.com/question/3450376#readmore



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Maurinko [17]

Answer:

a

Step-by-step explanation:

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2 years ago
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If a person tosses a coin 23 times, how many ways can he get 11 heads
EastWind [94]

Tossing a coin is a binomial experiment.

Now lets say there are 'n' repeated trials to get heads. Each of the trials can result in either a head or a tail.

All of these trials are independent since the result of one trial does not affect the result of the next trial.

Now, for 'n' repeated trials the total number of successes is given by

_{r}^{n}\textrm{C}

where 'r' denotes the number of successful results.

In our case n=23 and r=11,

Substituting the values we get,

_{11}^{23}\textrm{C}=\frac{23!}{11!\times 12!}

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3 0
2 years ago
What integer represents a rise in temperature of 19°?
nataly862011 [7]

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The opposite of that would be -19, read out as negative 19.

7 0
2 years ago
Candidate A is facing two opposing candidates in a mayoral election. In a recent poll of 300 residents, 98 supported candidate B
babymother [125]

Answer:

[0.4235, 0.5365]

Step-by-step explanation:

Data given and notation  

n=300 represent the random sample taken    

X=300-98-58=144 represent the people that support the candidate A in the sample

\hat p=\frac{144}{300}=0.48 estimated proportion of people that support the candidate A in the sample

\alpha=0.05 represent the significance level

Confidence =0.95 or 95%

p= population proportion of people that support the candidate A.

Confidence interval

The confidence interval would be given by this formula

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

For the 95% confidence interval the value of \alpha=1-0.95=0.05 and \alpha/2=0.025, with that value we can find the quantile required for the interval in the normal standard distribution.

z_{\alpha/2}=1.96

And replacing into the confidence interval formula we got:

0.48 - 1.96 \sqrt{\frac{0.48(1-0.48)}{300}}=0.4235

0.48 + 1.96 \sqrt{\frac{0.48(1-0.48)}{300}}=0.5365

And the 95% confidence interval would be given (0.4235;0.5365).

0.4235 \leq p \leq 0.5365

8 0
2 years ago
Chin researched the amount of money 150 students earned per month from jobs held during the summer. He created a table of six sa
dem82 [27]

the choices are below

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