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ehidna [41]
2 years ago
14

Write 81.42 as a mixed number

Mathematics
2 answers:
Kruka [31]2 years ago
7 0
81.42 =81 \frac{42}{100}=81\frac{21\times 2}{50\times 2} = \boxed{\bf{81\frac{21}{50}}}
Effectus [21]2 years ago
5 0
81.42
= 81+ 0.42
= 81+ 42/100
= 81+ (42/2) / (100/2)
= 81+ 21/50
= 81 21/50

The final answer is 81 21/50~
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If P = -5, and Q = -15 find (P + Q) x (P - Q)​
o-na [289]

Answer:

-200

Step-by-step explanation:

We are given that:

  • P= -5
  • Q= -15

Simply insert this values into their correspondents in the equation given.

(-5 + (-15)) * (-5 - (-15))

Subtracting two negatives is like adding a positive and a negative.

(-5 + (-15)) * (-5 + 15)

Solve inside the parenthesis.

(-20) * (10)

Multiply.

-200

5 0
2 years ago
Byron is conducting an experiment to determine whether a new medication is effective in reducing coughing. He finds 1,200 volunt
hammer [34]

Answer:

The medication is effective, but may take some time to work its way to the immune system

Step-by-step explanation:

3 0
2 years ago
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!

8 0
2 years ago
1.Hannah walked the same number of miles each day for 5 days. Let m represent the total number of miles Hannah walked. Write a l
mamaluj [8]

1) Given m represent the total number of miles Hannah walked.

Let us assume Hannah walked x miles each day.

So total number of miles Hannah walked in 5 days is m=5x.

2)Given Jimmy brought 2 books per day and Peter brought 4 books per day.

If d is the number of days the book drive, total number of books they brought

        =total number books brought by Jimmy+ total number of books brought by Peter                                                                          

        = 2d+4d = 6d

3) Given that c represent cost of the clothes.

Then 30% discount = 0.3c

Hence cost of the clothes = c-0.3c = 0.7c

4) Given c represent the cost of their meal before tipping.

And tip is 15% of the cost.

Hence tip = 0.15c

Hence total bill = c+0.15c=1.15c

5) Given that Tommy had 40$ in his pocket.

And pair of pants cost 37$ with 10% sales tax.

Hence cost of pants including sales tax = 37+37*0.1 = 40.1$ >40$

Hence Tommy cannot afford the pants with 40$ in his pocket.

For following problems, we should group the like terms together and then simplify.

6) (x+8)+(3x-12) = (x+3x)+8-12 = 4x-4

7) 5x+7-(3x-4) = 5x+7-3x+4 = (5x-3x)+7+4 = 2x+11

8) (x-2)-(-4x+1)=x-2+4x-1 = (x+4x)-2-1 = 5x-3

8 0
3 years ago
Complete the square to transform the expression x2 + 4x + 2 into the form a(x − h)2 + k.
Nina [5.8K]

Answer:

Step-by-step explanation:

y = (x^2 + 4x)      + 2

Take 1/2 of the linear term 4/2 = 2 and square that result. 2^2 = 4.

Put it after 4x

y = (x^2 + 4x + 4)   +2  Subtract what you put inside the brackets on the outside.

y = (x^2 + 4x + 4) + 2 - 4      Combine the right.

y = (x^2 + 4x + 4) - 2            Express the brackets as a square.

y = (x + 2)^2 - 2

That's your answer

a = 1

h = 2

k = -2

3 0
3 years ago
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