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dimulka [17.4K]
2 years ago
10

4. Sandra has 35% of the money she needs. She has $105. How much more money does she need? Write the equation and solve.​

Mathematics
1 answer:
guajiro [1.7K]2 years ago
5 0

Answer:

36.75

Step-by-step explanation:

First is turn the percent into a decimal which would be 0.35.

Then multiply 0.35 with 105

Finally this would equal to 36.75

Note: Hope I got it right, if I didn't I am very sorry this is all I can really do. If I did get it wrong please comment and I'll try again.

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Tina brought $37.75 to the art supply store. She bought a brush, a sketchbook, and a paint set. The brush was 1 6 as much as the
Firlakuza [10]

Answer:

18.75=p

s=14.06

b=2.34

Step-by-step explanation:

37.75=b+s+p

-4

33.75=b+s+p

s=3/4p

b=1/6s

33.75=1/6(3/4p)+3/4p+p

33.75=3/20p+3/4p+p

(33.75=18/20p+p)20

675=18p+p)/18

37.50=2p)/2

18.75=p

s=3/4p

3/4(18.75)

s=14.06

b=1/6s

1/6(14.06)

b=2.34

37.75-4=18.75+14.06+2.34=35.15

im missing 50 cents somewhere its wrong but close i think sorry

7 0
2 years ago
Juma earns $12.50 for each newspaper
Ulleksa [173]
400=12.50x+50

400-50=12.50x
350=12.50x
350÷12.50=x
28=x

he would have to sell 28 news paper subscriptions 
4 0
3 years ago
HELP ME PLEASE!!! ASAP THANK YOU SO MUCH
oksian1 [2.3K]

Answer:

Step-by-step explanation:

8 0
1 year ago
Express 0.000216 in scientific notation
damaskus [11]
2.16 x 10 to the power of -4
3 0
2 years ago
Data is collected to compare two different types of batteries. We measure the time to failure of 12 batteries of each type. Batt
-Dominant- [34]

Using the t-distribution, it is found that since the <u>test statistic is greater than the critical value for the right-tailed test</u>, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

At the null hypothesis, it is <u>tested if it does not outlast by more than 2 hours</u>, that is, the subtraction is not more than 2:

H_0: \mu_B - mu_A \leq 2

At the alternative hypothesis, it is <u>tested if it outlasts by more than 2 hours</u>, that is:

H_1: \mu_B - \mu_A > 2

  • The sample means are: \mu_A = 8.65, \mu_B = 11.23
  • The standard deviations for the samples are s_A = s_B = 0.67

Hence, the standard errors are:

s_{Ea} = S_{Eb} = \frac{0.67}{\sqrt{12}} = 0.1934

The distribution of the difference has <u>mean and standard deviation</u> given by:

\overline{x} = \mu_B - \mu_A = 11.23 - 8.65 = 2.58

s = \sqrt{s_{Ea}^2 + s_{Eb}^2} = \sqrt{0.1934^2 + 0.1934^2} = 0.2735

The test statistic is given by:

t = \frac{\overline{x} - \mu}{s}

In which \mu = 2 is the value tested at the hypothesis.

Hence:

t = \frac{\overline{x} - \mu}{s}

t = \frac{2.58 - 2}{0.2735}

t = 2.12

The critical value for a <u>right-tailed test</u>, as we are testing if the subtraction is greater than a value, with a <u>0.05 significance level</u> and 12 + 12 - 2 = <u>22 df</u> is given by t^{\ast} = 1.71

Since the <u>test statistic is greater than the critical value for the right-tailed test</u>, it is found that there is enough evidence to conclude that Battery B outlasts Battery A by more than 2 hours.

A similar problem is given at brainly.com/question/13873630

7 0
2 years ago
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