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3 years ago

The tasty Deli uses 6.5 lb of ham each day how many ounces of ham will be used in 3 weeks

Mathematics

2 answers:

vekshin13 years ago

8 0

136.5 lbs
There’s 16 ounces in a pound
135.5 • 16
Answer is 2,184 ounces of ham will be used in 3 weeks

yanalaym [24]3 years ago

5 0

Answer: 2184

Step-by-step explanation:

1lb = 16oz

6.5lb (16oz/1lb) = 104 oz

which means its 104 oz of ham uses in each day.

7 days in a week, then 3 weeks means 3 x 7 = 21 days

104oz in a day, 21 days (3 weeks) equals 21 x 104 = 2184 oz

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3 years ago

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A city planner makes a scale drawing of a proposed playground. The length of the actual playground is 78 feet, and the width is

Allushta [10]

Answer:

B) 6.5 inches by 3.5 inches

Step-by-step explanation:

Multiplying the actual dimensions by the scale gives the scale dimensions:

(0.5 in)/(6 ft) × {78 ft, 42 ft}

= {39/6 in, 21/6 in}

= {6.5 in, 3.5 in}

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Suppose that the length of a side of a cube X is uniformly distributed in the interval 9

Nastasia [14]

Answer:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

Step-by-step explanation:

Given

$9 < x < 10$ --- interval

Required

The probability density of the volume of the cube

The volume of a cube is:

$v = x^3$

For a uniform distribution, we have:

$x \to U(a,b)$

and

$f(x) = \left \{ {{\frac{1}{b-a}\ a \le x \le b} \atop {0\ elsewhere}} \right.$

$9 < x < 10$ implies that:

$(a,b) = (9,10)$

So, we have:

$f(x) = \left \{ {{\frac{1}{10-9}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Solve

$f(x) = \left \{ {{\frac{1}{1}\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$

Recall that:

$v = x^3$

Make x the subject

$x = v^\frac{1}{3}$

So, the cumulative density is:

$F(x) = P(x < v^\frac{1}{3})$

$f(x) = \left \{ {{1\ 9 \le x \le 10} \atop {0\ elsewhere}} \right.$ becomes

$f(x) = \left \{ {{1\ 9 \le x \le v^\frac{1}{3} - 9} \atop {0\ elsewhere}} \right.$

The CDF is:

$F(x) = \int\limits^{v^\frac{1}{3}}_9 1\ dx$

Integrate

$F(x) = [v]\limits^{v^\frac{1}{3}}_9$

Expand

$F(x) = v^\frac{1}{3} - 9$

The density function of the volume F(v) is:

$F(v) = F'(x)$

Differentiate F(x) to give:

$F(x) = v^\frac{1}{3} - 9$

$F'(x) = \frac{1}{3}v^{\frac{1}{3}-1}$

$F'(x) = \frac{1}{3}v^{-\frac{2}{3}}$

$F(v) = \frac{1}{3}v^{-\frac{2}{3}}$

So:

$f(v) = \left \{ {{\frac{1}{3}v^{-\frac{2}{3}}\ 9^3 \le v \le 10^3} \atop {0, elsewhere}} \right.$

8 0

3 years ago

The tasty Deli uses 6.5 lb of ham each day how many ounces of ham will be used in 3 weeks ​

The tasty Deli uses 6.5 lb of ham each day how many ounces of ham will be used in 3 weeks