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4 years ago

10

Ebony is building a shelf to hold the two boxes shown. What is the least width she should make the shelf?

1 answer:

ki77a [65]4 years ago

7 0

There exists the same question from other source that shows the two boxes.

Box 1 has a length of 4/5 feet
Box 2 has a length of 3/4 feet

Because Ebony is trying to find the total amount of space needed to hold the boxes, use addition.

Here, is the solution:
= 4 / 5 + 3 / 4
= (16 + 15) / 20
= 31 / 20

So, Ebony can make a shelf at least 31/20 feet wide or 1 11/20 feet.

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From 1900 to 2000 the average temperature in July in Atlanta was 90∘F. A researcher believes due to Global Warming that the aver

ElenaW [278]

Answer:

Option C

Step-by-step explanation:

Hypothesis is a tentative guess which can be statistically proven to be valid or invalid depending with the given level of significance. Normally, we have null and alternative hypothesis. For this question, the null hypothesis is that the mean temperature in Atlanta in July is less than or equal to $90^{\circ} F$

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3 years ago

Solve 15-y>8 show working

Mashcka [7]

Answer:

y <7

Step-by-step explanation:

15-y>8

Subtract 15 from each side

15-15-y>8-15

-y>-7

Multiply each side by -1, remembering to flip the inequality

-1 * -y > -7*-7

y <7

6 0

3 years ago

Which formula can be used to describe the sequence <br> -3, 3/5, -3/25, 3/125, -3/625

maks197457 [2]

Answer:

a(n) = -3(-1/5)^(n - 1)

Step-by-step explanation:

From the first term, -3, we get the second term, 3/5, by multiplying -3 by -1/5.

Thus, the common ratio is -1/5.

The general formula in this case is a(n)=(first term)(common ratio)^(n - 1), or

a(n) = -3(-1/5)^(n - 1)

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3 years ago

Convert the following repeating decimal into a fraction<br> _<br> 0.497

Annette [7]

The fraction is 497/1000

6 0

3 years ago

In a large school, it was found that 77% of students are taking a math class, 74% of student are taking an English class, and 70

Iteru [2.4K]

Answer:

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

Step-by-step explanation:

We solve this question working with the probabilities as Venn sets.

I am going to say that:

Event A: Taking a math class.

Event B: Taking an English class.

77% of students are taking a math class

This means that $P(A) = 0.77$

74% of student are taking an English class

This means that $P(B) = 0.74$

70% of students are taking both

This means that $P(A \cap B) = 0.7$

Find the probability that a randomly selected student is taking a math class or an English class.

This is $P(A \cup B)$ , which is given by:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

So

$P(A \cup B) = 0.77 + 0.74 - 0.7 = 0.81$

0.81 = 81% probability that a randomly selected student is taking a math class or an English class.

Find the probability that a randomly selected student is taking neither a math class nor an English class.

This is

$1 - P(A \cup B) = 1 - 0.81 = 0.19$

0.19 = 19% probability that a randomly selected student is taking neither a math class nor an English class

6 0

3 years ago