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emmainna [20.7K]
2 years ago
10

How do you write 3 8 / 9 as a decimal?

Mathematics
2 answers:
MAXImum [283]2 years ago
5 0

Answer:

3.8888888888889

Step-by-step explanation:

Sliva [168]2 years ago
3 0

Answer:

3.888

Step-by-step explanation:

Turn 3 8/9 into a mixed number

3 8/9 = 39/9

Simplify 39/9 and you'll get 3.888

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Do the following lengths satisfy the Pythagorean Theorem? Lengths =14, 50, 48.<br> Yes<br> No
seraphim [82]

Answer:

  yes

Step-by-step explanation:

  14² +48² = 196 +2304 = 2500 = 50²

The three lengths (14, 48, 50) are a Pythagorean triple.

__

<em>Additional comment</em>

The two smaller numbers correspond to side lengths. The larger one corresponds to the hypotenuse length.

We typically write the numbers of a Pythagorean triple in increasing order, but the order doesn't really matter. As sides of a triangle, the triangle formed is a right triangle, regardless of the order.

6 0
2 years ago
Johnson rode her bike 7 miles to a lake. She rode 3/4 of distance to the lake before lunch. How many miles did Johnson ride befo
Virty [35]

Answer:

5.25 miles.

Step-by-step explanation:

What I did was I took 3/4 and made it into its decimal form which is 3/4. I then times it by 7 to get 5.25.

5.25 miles is your answer.

5 0
3 years ago
Read 2 more answers
Exercise 11.21) Many smartphones, especially those of the LTE-enabled persuasion, have earned a bad rap for exceptionally bad ba
stiv31 [10]

Answer:

Step-by-step explanation:

Hello!

The researcher suspects that the battery life between charges for the Motorola Droid Razr Max differs if its primary use is talking or if its primary use is for internet applications.

Since the means for talk time usage (20hs) is greater than the mean for internet usage (7hs) the main question is if the variance in hours of usage is also greater when the primary use is talk time.

Be:

X₁: Battery duration between charges when the primary usage of the phone is talking. (hs)

n₁= 12

X[bar]₁= 20.50 hs

S₁²= 199.76hs² (S₁= 14.13hs)

X₂: Battery duration between charges when the primary usage of the phone is internet applications.

n₂= 10

X[bar]₂= 8.50

S₂²= 33.29hs² (S₂= 5.77hs)

Assuming that both variables have a normal distribution X₁~N(μ₁;σ₁²) and X₂~N(μ₂; σ₂²)

The parameters of interest are σ₁² and σ₂²

a) They want to test if the population variance of the duration time of the battery when the primary usage is for talking is greater than the population variance of the duration time of the battery when the primary usage is for internet applications. Symbolically: σ₁² > σ₂² or since the test to do is a variance ratio: σ₁²/σ₂² > 1

The hypotheses are:

H₀: σ₁²/σ₂² ≤ 1

H₁: σ₁²/σ₂² > 1

There is no level of significance listed so I've chosen α: 0.05

b) I've already calculated the sample standard deviations using a software, just in case I'll show you how to calculate them by hand:

S²= \frac{1}{n-1}*[∑X²-(∑X)²/n]

For the first sample:

n₁= 12; ∑X₁= 246; ∑X₁²= 7240.36

S₁²= \frac{1}{11}*[7240.36-(246)²/12]= 199.76hs²

S₁=√S₁²=√199.76= 14.1336 ≅ 14.13hs

For the second sample:

n₂= 10; ∑X₂= 85; ∑X₂²= 1022.12

S₂²= \frac{1}{9}*[1022.12-(85)²/10]= 33.2911hs²

S₂=√S₂²=√33.2911= 5.7698 ≅ 5.77hs

c)

For this hypothesis test, the statistic to use is a Snedecors F:

F= \frac{S_1^2}{S_2^2} *\frac{Sigma_1^2}{Sigma_2^2} ~~F_{n_1-1;n_2-1}

This test is one-tailed right, wich means that you'll reject the null hypothesis to big values of F:

F_{n_1-1;n_2-1; 1-\alpha }= F_{11;9;0.95}= 3.10

The rejection region is then F ≥ 3.10

F_{H_0}= \frac{199.76}{33.29} * 1= 6.0006

p-value: 0.006

Considering that the p-value is less than the level of significance, the decision is to reject the null hypothesis.

Then at a 5% level, there is significant evidence to conclude that the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for talk is greater than the population variance of the duration time of the batteries of Motorola Droid Razr Max smartphones used primary for internet applications.

I hope this helps!

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