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Blababa [14]
3 years ago
9

Margaret is cutting a block of cheese into smaller cubes measuring 1 cm by 1 cm by 1 cm.How many unit cubes of cheese can she cu

t from the large block?
Mathematics
1 answer:
nirvana33 [79]3 years ago
8 0
How what size is the large block?
Convert the measurements of the large block to cm.
Then calculate the Volume of the large cube.
The answer will be in cubic cm. That is the number of 1 cm cubes.
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X + 2y = 2<br><br>A, B, C, or D​
dedylja [7]

Answer:

Graph D

Step-by-step explanation:

Convert to y-intercept:

x + 2y = 2

2y = -x + 2

y = -1/2x + 1

Slope = -1/2 (1 down, 2 right)

y-intercept = + 1

5 0
3 years ago
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Seven friends go to the local sandwich shop. All six people order the combination meal, which includes a drink and a veggie wrap
galben [10]

Answer: B

Step-by-step explanation:

$4.95 x 6 people = 29.7 ≈ $30

4 0
1 year ago
A group of hikers buys 8 bags of mixed nuts. Each bag of nuts contains 3 1/2 ounces of mixed nuts. There are 12 hikers on the tr
Travka [436]

Multiply the number of bags by the ounces in each bag:

8 x 3 1/2 = 28 total ounces.

Divide total ounces by the number of hikers:

28 / 12 = 2.33 ounces per hiker.

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Divide the number of ounces per hiker by the number of ounces in 1 cup:

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4 0
3 years ago
If g(x) = 3x + 8, evaluate g(2)
Mkey [24]

Answer:

g(2) = 14

Step-by-step explanation:

g(x) = 3x + 8

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g(2)=14 Hope this helps

7 0
3 years ago
In a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50. Determ
shepuryov [24]

Answer:

t-distribution should be used to construct a confidence interval.

Step-by-step explanation:

We are given that a random sample of 18 families, the average weekly food expense was $95.60 with a sample standard deviation of $22.50.

We have to determine whether a normal distribution (Z values) or a t- distribution should be used or whether neither of these can be used to construct a confidence interval.

<em><u>Since in this question we are provided with;</u></em>

Sample average weekly food expense, \bar X = $95.60

Sample standard deviation, s = $22.50

Sample of families, n = 18

The distribution that we will use here to construct a confidence interval will be <u>t-distribution</u> because in the question we don't know anything about population standard deviation (\sigma) .

Normal distribution is used when we know population standard deviation (\sigma).

So, the pivotal quantity for confidence interval that will be used is One-sample t-test statistics;

                P.Q. = \frac{\bar X -\mu}{\frac{s}{\sqrt{n} } } ~ t_n_-_1

Therefore, t-distribution should be used to construct a confidence interval.

3 0
3 years ago
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