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Alex_Xolod [135]
3 years ago
10

Tara has 4 4/5 feet of rope. Part A: How many 2/5 foot pieces can Tara cut from the 4 4/5 feet of rope? Show your work. Part B:

Using the information in Part A, interpret the meaning of the quotient in terms of the two fractions given.
Mathematics
1 answer:
aleksley [76]3 years ago
3 0

Answer:

Part A

12 pieces

Part B

Tara can cut Twelve 2/5 foot pieces can be cut from 4 4/5 feet of rope.

Step-by-step explanation:

Part A: How many 2/5 foot pieces can Tara cut from the 4 4/5 feet of rope?

This is calculated as:

4 4/5 feet of rope ÷ 2/5 foot pieces

= 24/5 ÷ 2/5

= 24/5 × 5/2

= 12

Part B: Using the information in Part A, interpret the meaning of the quotient in terms of the two fractions given

The quotient in Part A is 12

Therefore, this can be interpreted as:

Tara can cut Twelve 2/5 foot pieces can be cut from 4 4/5 feet of rope.

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Which equation could possibly represent the graphed function?
Veseljchak [2.6K]

Answer:

A

Step-by-step explanation:

You're looking for x-intercepts

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x = -4, x = -2, x = 4

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so you can rewrite each x-intercept as an equation

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The top of a 230 foot vertical tower is to be anchored by cables that make an angle with a measure of 50 degrees with the ground
Triss [41]

9514 1404 393

Answer:

  • cable length: 300.2 ft
  • anchor distance: 193.0 ft

Step-by-step explanation:

The side given is opposite the given angle. We want to find both the hypotenuse and the adjacent side in the right triangle that models the geometry.

  Sin = Opposite/Hypotenuse

  sin(50°) = (230 ft)/h

  h = (230 ft)/sin(50°) ≈ 300.24 ft

The length of the cable must be about 300.2 feet.

__

  Tan = Opposite/Adjacent

  tan(50°) = (230 ft)/d

  d = (230 ft)/tan(50°) ≈ 192.99 ft

The cable must be anchored about 193.0 ft from the tower.

4 0
3 years ago
It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per nig
Kitty [74]

Answer: No , at 0.05 level of significance , we have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

Step-by-step explanation:

Let \mu denotes the average hours of sleep per night.

As per given , we have

H_0:\mu=7\\H_a:\mu

, since H_a is left-tailed and population standard deviation is unknown, so the test is a left-tailed t -test.

Also , it is given that ,

Sample size : n= 22

Sample mean : \overline{x}=7.24

Sample  standard deviation : s= 1.93

Test statistic : t=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}

i.e.  t=\dfrac{7.24-7}{\dfrac{1.93}{\sqrt{22}}}\approx0.58

For significance level \alpha=0.05 and degree of freedom 21 (df=n-1),

Critical t-value for left-tailed test= t_{\alpha, df}=t_{0.05,21}=- 1.7207

Decision : Since the test statistic value (0.58) > critical value  1.7207, it means we are failed to reject the null hypothesis .

[Note : When |t_{cal}|>|t_{cri}|, then we accept the null hypothesis.]

Conclusion: We have sufficient evidence to reject the claim that LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average.

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