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ziro4ka [17]
3 years ago
13

Natalie went to the shoe store and bought two pairs of sneakers. One pair was $35.60, and the other was $49.95. If she has $100.

00, how much will she have left after buying sneakers?
Mathematics
2 answers:
makkiz [27]3 years ago
8 0
She will have 14.45 left over
Contact [7]3 years ago
6 0

First you need to add the prices to get 85.55. Then 100 - 85.55 is 14.45.




She will have $14.45 leftover

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Determine the number of solutions that exist to the equation below 8(j-4)=2(4j-16)
Firdavs [7]
Hello :
<span>8(j-4)=2(4j-16)
</span><span>2(4j-16)= 2(4(j-4))=8(j-4)
</span>8(j-4)= 8(j-4)....(identity : infinty solutions)

5 0
3 years ago
What is <br> r squared+7r+6=0
Firlakuza [10]

Answer:

r = -6, -1

Step-by-step explanation:

r^2 + 7r+6 =0

Factor

What 2 numbers multiply to 6 and add to 7

6*1 =6

6+1 =7

(r+6) (r+1) =0

Using the zero product property

r+6 =0        r+1 =0

r = -6            r=-1

7 0
3 years ago
Ratio, Rate, and Speed
Andre45 [30]

Answer:

10:12:15

Step-by-step explanation:

a/b=5/6

b/c=4/5

thus

c/b=5/4

LCM of 4 and 6=12

5/6=10/12

5/4=15/12

b is the 12 in both

thus ratio between a:b:c is 10:12:15

i can reply in comments if more explanation is needed

8 0
3 years ago
7b+12=5b what is the value for b?
likoan [24]

Answer:

B = -6

Step-by-step explanation:

7b + 12 = 5b

12 = -2b

-6 = b

Hope this helps. Pls give brainliest.

6 0
2 years ago
A college conducts a common test for all the students. For the Mathematics portion of this test, the scores are normally distrib
Jet001 [13]

Using the normal distribution, it is found that 58.97% of students would be expected to score between 400 and 590.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

Z = \frac{X - \mu}{\sigma}

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 502, \sigma = 115

The proportion of students between 400 and 590 is the <u>p-value of Z when X = 590 subtracted by the p-value of Z when X = 400</u>, hence:

X = 590:

Z = \frac{X - \mu}{\sigma}

Z = \frac{590 - 502}{115}

Z = 0.76

Z = 0.76 has a p-value of 0.7764.

X = 400:

Z = \frac{X - \mu}{\sigma}

Z = \frac{400 - 502}{115}

Z = -0.89

Z = -0.89 has a p-value of 0.1867.

0.7764 - 0.1867 = 0.5897 = 58.97%.

58.97% of students would be expected to score between 400 and 590.

More can be learned about the normal distribution at brainly.com/question/27643290

#SPJ1

6 0
2 years ago
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