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Veronika [31]
3 years ago
7

If 5 cables cost $75, what is the cost of 1 cable?

Mathematics
2 answers:
Dimas [21]3 years ago
6 0
The answer is 15, 75 divided by 5 would be 15 so 1 cable is $15
mylen [45]3 years ago
5 0

Answer:

$15

Step-by-step explanation:

75/5=15

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What is the equivalent fraction of 21/10 with denominater 10 (pls help its urgent also then i will mark brainlist)
mart [117]

Answer: u can research it up

Step-by-step explanation: 210/100 i guess

8 0
2 years ago
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Solve these problems and show your work! ONLY HAND WRITTEN answers will be accepted! When
Dahasolnce [82]

For Kohl’s:

Purchase Price:

$60

Discount:

(60 x 25)/100 = $15.00

Final Price:

60 - 15.00 = $45.00

You would save $15.00 from Kohl’s and pay only $45.00

For Target:

Purchase Price:

$38

Discount:

(38 x 15)/100 = $5.70

Final Price:

38 - 5.70 = $32.30

You would save $5.70 from Target and pay only $32.30

Hope this helps! :)

3 0
3 years ago
Given f(x) and g(x) = f(x) + k, look at the graph below and determine the value of k.
Anton [14]

Answer:

Given the graph f(x) = \frac{1}{3}x -2 and g(x) = \frac{1}{3}x + 3

We have to find the value of k;

Since, g(x) = f(x) +k

\frac{1}{3}x+3 = \frac{1}{3}x-2 +k

Subtract \frac{1}{3}x from both sides we get;

3 = -2 +k

Add 2 to both sides we get;

3+2 = -2 +k +2

Simplify:

5 = k

or

k = 5

Therefore, the value of k = 5


3 0
3 years ago
Read 2 more answers
Which expression is equivalent to the following complex fraction? <br>​
Sladkaya [172]

Answer:

Option B

Step-by-step explanation:

Answered by Gauthmath

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5 0
3 years ago
A survey of 542 consumers reveals that 301 favor the new design for a product. Construct a 90% confidence interval for the true
kodGreya [7K]

Answer:

The 90% confidence interval for the true proportion of all consumers who favor the design is (0.52, 0.59).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 542, \pi = \frac{301}{542} = 0.555

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.555 - 1.645\sqrt{\frac{0.555*0.445}{542}} = 0.52

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.555 + 1.645\sqrt{\frac{0.555*0.445}{542}} = 0.59

The 90% confidence interval for the true proportion of all consumers who favor the design is (0.52, 0.59).

5 0
3 years ago
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