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3 years ago

Ann works in a supermarket for $15.00 per hour. If her pay is increased to $18.00, then what is her percent increase in pay?

Mathematics

2 answers:

wariber [46]3 years ago

4 0

Answer:

previous salary of Ann=$15

Present salary of Ann=$18

The increase in salary=$18-$15

=$3

%increase in salary=increase in salary/previous salary

=($3/$15)×100%

=20%

vfiekz [6]3 years ago

4 0

She would increase 20% of her normal pay

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lesantik [10]

Answer:

$7500

Step-by-step explanation:

$5000 was deposited

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so in ten years the interest will be 10 ×$250 =$2500

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3 years ago

I need help with #65 to #70 ASAP please and thank you …. Could someone please help me with it… I need to get it done ASAP… it is

never [62]

All of the given questions(#65 to #70) are AP(s).

Formula to be used: a(n) = a(1) + (n - 1)d

Pronounced as:

nth term = 1st term + (n - 1)d, where d is the common difference.

*d = difference of any two consecutive terms.

65):

1st term = 2, d = 6 - 2 = 4

Therefore,

11th term = 2 + (11 - 1)4 = 42

66):

1st term = 5, d = 15 - 5 = 10

Therefore,

16th term = 5 + (16 - 1)10 = 155

67):

1st term = 19, d = 39 - 19 = 20

Therefore,

21st term = 19 + (21 - 1)20 = 419

68):

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36th term = 8 + (36 - 1)30 = 1058

69):

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70):

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3 years ago

A study sampled 350 upperclassmen (Group 1) and 250 underclassmen (Group 2) at high schools around the city of Houston. The stud

iris [78.8K]

Answer:

$-0.048$

Do not reject $H_0:P_1-P_2=0$

Step-by-step explanation:

From the question we are told that

Sample size $n_1=350$

Sample size $n_2=250$

Sample proportion 1 $\hat P= \frac{25}{350} =>0.07$

Sample proportion 2 $\hat P= \frac{19}{250} =>0.076$

95% confidence interval

Generally for 95% confidence level

Level of significance

$\alpha = 1-0.95=>0.05$

$\alpha /2=\frac{0.05}{2} =>0.025$

Therefore

$Z_a_/_2=1.96$

Generally the equation for confidence interval between $P_1 - P_2$ is mathematically given as

$(\hat P_1-\hat P_2)\pm Z_a_/_2\sqrt{\frac{\hat P_1(1-\hat P_1)}{n_1}+\frac{\hat P_2(1-\hat P_2)}{n_2} }$

$(0.07-0.076)\pm 1.96\sqrt{\frac{0.07(1-0.07)}{350}+\frac{0.076(1-0.076)}{250} }$

$(0.07-0.076)\pm 1.96\sqrt{4.66896*10^-^4 }$

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$(-0.006)- 0.042=>-0.048$

$(-0.006)+ 0.042=>0.036$

Therefore

Confidence interval is

$-0.048$

Conclusion

Given the confidence interval has zero

Therefore do not reject $H_0:P_1-P_2=0$

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2 years ago

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sertanlavr [38]

Answer:

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3 years ago

Ann works in a supermarket for $15.00 per hour. If her pay is increased to $18.00, then what is her percent increase in pay?​

Ann works in a supermarket for $15.00 per hour. If her pay is increased to $18.00, then what is her percent increase in pay?