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3 years ago

Bonita said that the product of 5/6 times 1 2/3= 7/3. How can you tell that her answer is wrong?

Mathematics

1 answer:

Masja [62]3 years ago

6 0

Answer:

B. 5/6 of a number cannot be greater than the number.

7/3 = 2 1/3 > 1 2/3

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The perpendicular line of x-6y=2, (2, 4) in slope intercept form

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The perpendicular line to x-6y=2, and passing through (2, 4) is y=-6x+16

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3 years ago

Solve the system of equations using the substitution method. Show your work and be sure to include the solution to the system y=

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Is that simultaneous equation

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2 years ago

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A caterpillar eats through 2/3 leaves for every 4/5 minutes. What is the unit ratio for leaves eaten through to minutes?

eimsori [14]

Answer:

1 1/5

Step-by-step explanation:

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3 years ago

Charlie does a weekly exercise program consisting of cardiovascular work and weight training. Each week, he exercises for at lea

german

Answer:

$(x_1,y_1) = (0,9)$

$(x_2,y_2) = (6,3)$

$(x_3,y_3) = (6,12)$

Step-by-step explanation:

Given

x = hours spent on cardiovascular work

y = hours spent on weight training

From the first statement,

Each week: he spends at least 9 hours

This is represented as:

$Total \ge 9$

$x + y \ge 9$

From the second statement,

Each week: he spends a max of 12 hours on weights

This is represented as:

$Weight \le 12$

$y \le 12$

From the third statement,

Each week: he spends a max of 6 hours on cardio works

This is represented as:

$Cardio \le 6$

$x \le 6$

So, the inequalities are:

$x + y \ge 9$

$y \le 12$

$x \le 6$

See attachment for the graph where the solution are also highlighted

From the attachment, the solutions are:

$(x_1,y_1) = (0,9)$

$(x_2,y_2) = (6,3)$

$(x_3,y_3) = (6,12)$

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3 years ago

According to the article "Fatigue Testing of Condoms" (Polymer Testing, 2009: 567–571), "tests currently used for condoms are su

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Answer:

99% Confidence interval: (1196,1973)

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 20

Mean, μ = 1584

Standard Deviation, σ = 607

99% Confidence interval:

$\bar{x} \pm t_{critical}\displaystyle\frac{s}{\sqrt{n}}$

Putting the values, we get,

$t_{critical}\text{ at degree of freedom 19 and}~\alpha_{0.01} = \pm 2.86$

$1584 \pm 2.86(\dfrac{607}{\sqrt{20}} )\\\\ = 1584 \pm 388.18\\ = (1195.82 ,1972.18)\\ \approx (1196,1973)$

Thus, there is a 99% chance that it will break in approximately 1196 to 1972

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3 years ago