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netineya [11]
3 years ago
6

Kate is allowed to work no more than 20 hours a week. She has already worked 13 hours this week. At most, how many more hours CA

N she work? Write an inequality and solve.
A) x + 13 ≤ 20; x ≤ 7
B) x + 13 ≥ 20; x ≥ 7
C) x + 13 < 20; x < 7
D) x + 13 > 20; x > 7
Mathematics
1 answer:
Alja [10]3 years ago
8 0

Answer:

A) \displaystyle x + 13 ≤ 20; x ≤ 7

Step-by-step explanation:

The keywords are no more than, meaning that twenty hours is the MAXIMUM, or LIMIT, so you would choose this answer choice.

* There IS a range, so you must include <em>equal</em><em> </em><em>to</em>.

I am joyous to assist you anytime.

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Answer:

B. 1/m¹⁸

Step-by-step explanation:

To simplify the equation, we start with the values inside the brackets.

Therefore  m⁻¹m⁵ results to m⁴.

This is because the sign between the m⁻¹ and m⁵ is multiplication, and when multiplying figures that have a similar base, we add the indices.

that is, -1+5=4

then we divide m⁴ by m⁻², that is, m⁴/m⁻²=m⁶

To divide figures that have the same base we subtract the powers.

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3 years ago
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Grace gets<br> paid $5 per<br> hour to<br> babysit her<br> younger<br> siblings.
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Step-by-step explanation:

You must assume Grace has 0 dollars.

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3 years ago
The amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5. C
Bingel [31]

Answer:

The required 97.5% confidence interval is

\text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\\text {CI} = 15.5 \pm 2.8412\cdot \frac{0.31}{\sqrt{8} } \\\\\text {CI} = 15.5 \pm 2.8412\cdot 0.1096\\\\\text {CI} = 15.5 \pm  0.311\\\\\text {CI} = 15.5 - 0.311, \: 15.5 + 0.311\\\\\text {CI} = (15.19, \: 15.81)\\\\

Therefore, we are 97.5% confident that the actual mean amount of juice in all such bottles is within the range of 15.19 to 15.81 ounces

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Step-by-step explanation:

The amounts (in ounces) of juice in eight randomly selected juice bottles are:

15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5

Let us first compute the mean and standard deviation of the given data.

Using Excel,

=AVERAGE(number1, number2,....)

The mean is found to be

\bar{x} = 15.5

=STDEV(number1, number2,....)

The standard deviation is found to be

s = 0.31  

The confidence interval for the mean amount of juice in all such bottles is given by

$ \text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $\\\\

Where \bar{x} is the sample mean, n is the samplesize, s is the sample standard deviation and t_{\alpha/2} is the t-score corresponding to a 97.5% confidence level.

The t-score corresponding to a 97.5% confidence level is

Significance level = α = 1 - 0.975 = 0.025/2 = 0.0125

Degree of freedom = n - 1 = 8 - 1 = 7

From the t-table at α = 0.0125 and DoF = 7

t-score = 2.8412

So the required 97.5% confidence interval is

\text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\\text {CI} = 15.5 \pm 2.8412\cdot \frac{0.31}{\sqrt{8} } \\\\\text {CI} = 15.5 \pm 2.8412\cdot 0.1096\\\\\text {CI} = 15.5 \pm  0.311\\\\\text {CI} = 15.5 - 0.311, \: 15.5 + 0.311\\\\\text {CI} = (15.19, \: 15.81)\\\\

Therefore, we are 97.5% confident that the actual mean amount of juice in all such bottles is within the range of 15.19 to 15.81 ounces.

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A line segment has endpoints at (3, 2) and (2, –3). Which reflection will produce an image with endpoints at (3, –2) and (2, 3)?
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For this case we have the following points:

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We observe that the points obtained are the points that are sought.

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