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Sindrei [870]
3 years ago
11

Need help with this. Hope someone can help

Mathematics
1 answer:
kirill115 [55]3 years ago
4 0

To solve this problem, we must remember that the formula for volume is V = lwh, where l represents length, w represents width, and h represents height. Since we know the respective values for length, height, and volume, we can substitute those into the formula and then solve for the width, as modeled below:

30,000 cm³ = 30 cm * w * 500 mm

We must recognize first, though, that one of the values is given in millimeters, not centimeters. To continue solving this problem, we must convert this value to centimeters, using our knowledge that 1 cm = 10mm. This means that 500 mm = 50 cm, so we can substitute this into our equation below:

30,000 cm³ = 30 cm * w * 50 cm

To begin to solve this equation, we should first multiply all of the factors together on the right side of the equation, which gives us:

30,000 cm³ = 1,500w cm²

Next, we must divide both sides of the equation by 1,500 in order to get the variable w alone on the right side of the equation.

20 cm = w

Therefore, the width of the figure shown above is 20 cm.

Hope this helps!

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svetoff [14.1K]

Answer: 23

Step-by-step explanation:

6 0
3 years ago
Tell weather the two rates form a proportion. help
Sindrei [870]
7/9 = 42/54...when 42/54 reduces, it becomes 7/9...so this is proportional because proportions are nothing but equivalent fractions.

u could also check it this way...
12/21 = 15/24
cross multiply
(21)(15) = (12)(24)
315 = 288.....these do not equal and are therefore, not proportionate

440/4 = 300/3
110 = 100....not equal, so not proportionate

120/5 = 88/4
24 = 22.....not proportionate

66/82 = 99/123
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6 0
3 years ago
Suppose that the distribution of typing speed in words per minute (wpm) for experienced typists using a new type of split keyboa
KengaRu [80]

Answer:

a. <u>0.5 or 50%</u>

b. <u>0.496 or 49.6%</u>

c. <u>0.8185 or 81.85%</u>

d. <u>Yes, it would be just 0.0013 or 0.13% of probability to find a typist whose speed exceeds 105 wpm</u>

e. <u>0.0252 or 2.52%</u>

f. <u>The qualifying speed would be 47.4 wpm.</u>

Step-by-step explanation:

a. Let's find the z-score this way:

μ  = 60 σ= 15

z-score = (x - μ)/σ

z-score = (60 - 60)/15

z-score = 0

Now, let's calculate the p value for z-score = 0, using the z-table:

<u>p (z=0) = 0.5 or 50%</u>

b. z-score = (x - μ)/σ

z-score = (59.9 - 60)/15

z-score = -0.01

Now, let's calculate the p value for z-score = -.0.01, using the z-table:

<u>p (z = -0.01) = 0.496 or 49.6%</u>

If the question were less or equal than 60, a and b would have the same answer. But in this case, the question is "less than 60 wpm".

c.

z-score = (x - μ)/σ

z-score = (45 - 60)/15

z-score = - 1

Now, let's calculate the p value for z-score = -1, using the z-table:

p (z = -1) = 0.1587

z-score = (x - μ)/σ

z-score = (90 - 60)/15

z-score = 2

Now, let's calculate the p value for z-score = 2, using the z-table:

p (z = 2) = 0.9772

<u>In consequence,</u>

<u>p (-1 ≤ z ≤ 2) = 0.9772 - 0.1587 = 0.8185 or 81.85%</u>

d. z-score = (x - μ)/σ

z-score = (105 - 60)/15

z-score = 3

Now, let's calculate the p value for z-score = 3, using the z-table:

p (z = 3) = 0.9987

In consequence,

p (z > 3) = 1 - 0.9987 = 0.0013

<u>Yes, it would be just a 0.13% of probability to find a typist whose speed exceeds 105 wpm.</u>

e. z-score = (x - μ)/σ

z-score = (75 - 60)/15

z-score = 1

Now, let's calculate the p value for z-score = 1, using the z-table:

p (z = 1) = 0.8413

In consequence,

p (z > 1) = 1 - 0.8413 = 0.1587

and if the two typists are independently selected, then

p = 0.1587 * 0.1587

<u>p = 0.0252 or 2.52%</u>

f. p = 0.2, using the z-table, the z-score is -0.84, then:

z-score = (x - μ)/σ

-0.84 = (x - 60)/15

-12.6 = x - 60

-x = -60 + 12.6

-x = - 47.4

x = 47.4

<u>The qualifying speed would be 47.4 wpm.</u>

8 0
3 years ago
Can anyone help me please​
MaRussiya [10]

Answer:

No it's not prism

Step-by-step explanation:

Because it has curved side. A prism has flat side not curved side

4 0
3 years ago
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