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navik [9.2K]
3 years ago
15

True/False: 2(10x - 5) + 8 is an expression.

Mathematics
1 answer:
natka813 [3]3 years ago
8 0

Answer:

True

Step-by-step explanation:

Itis true because you don't have an equal sign. Expressions don't have an equal sign, equations have an equal sign.

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Can anyone help me please? I have no idea what to do! :'(
liberstina [14]

Answer:

7) 10^(3/2)

8) 2^(1/6)

9) 2^(5/4)

10) 5^(5/4)

Step-by-step explanation:

7)  (√10)^3 = 10^(3/2)

8)  6 root 2 = 2^(1/6)

9)  (4 root 2)^5 = 2^(5/4)

10) (4 root 5)^5 = 5^(5/4)

5 0
3 years ago
Y=x+1<br> y+x=5<br> Cant figure out what i am suppose to do with this problem
exis [7]

Answer:can you give me more detail

Step-by-step explanation:

8 0
3 years ago
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A random sample of 49 lunch customers was taken at a restaurant. The average amount of time the customers in the sample stayed i
Hunter-Best [27]

Answer:

a)  σ/√n= 1.43 min

c) Margin of error 2.8028min

d) [30.1972; 35.8028]min

e) n=62 customers

Step-by-step explanation:

Hello!

The variable of interest is

X: Time a customer stays at a restaurant. (min)

A sample of 49 lunch customers was taken at a restaurant obtaining

X[bar]= 33 mi

The population standard deviation is known to be δ= 10min

a) and b)

There is no information about the distribution of the population, but we know that if the sample is large enough, n≥30, we can apply the central limit theorem and approximate the distribution of the sample mean to normal:

X[bar]≈N(μ;σ²/n)

Where μ is the population mean and σ²/n is the population variance of the sampling distribution.

The standard deviation of the mean is the square root of its variance:

√(σ²/n)= σ/√n= 10/√49= 10/7= 1.428≅ 1.43min

c)

The CI for the population mean has the general structure "Point estimator" ± "Margin of error"

Considering that we approximated the sampling distribution to normal and the standard deviation is known, the statistic to use to estimate the population mean is Z= (X[bar]-μ)/(σ/√n)≈N(0;1)

The formula for the interval is:

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

The margin of error of the 95% interval is:

Z_{1-\alpha /2}= Z_{1-0.025}= Z_{0.975}= 1.96

d= Z_{1-\alpha /2}*(σ/√n)= 1.96* 1.43= 2.8028

d)

[X[bar]±Z_{1-\alpha /2}*(σ/√n)]

[33±2.8028]

[30.1972; 35.8028]min

Using a confidence level of 95% you'd expect that the interval [30.1972; 35.8028]min contains the true average of time the customers spend at the restaurant.

e)

Considering the margin of error d=2.5min and the confidence level 95% you have to calculate the corresponding sample size to estimate the population mean. To do so you have to clear the value of n from the expression:

d= Z_{1-\alpha /2}*(σ/√n)

\frac{d}{Z_{1-\alpha /2}}= σ/√n

√n*(\frac{d}{Z_{1-\alpha /2}})= σ

√n= σ* (\frac{Z_{1-\alpha /2}}{d})

n=( σ* (\frac{Z_{1-\alpha /2}}{d}))²

n= (10*\frac{1.96}{2.5})²= 61.47≅ 62 customers

I hope this helps!

3 0
3 years ago
CALLING ALL EXPERTS, PLEASE HELP ME WITH MY HOMEWORK!! WORTH A LOT OF POINTS AND WILL GIVE BRAINLIEST!!
laiz [17]

Answer:

1= 1 7/8 or 1.875

2= 235.75

3= 10 1/2 or 10.5

4=191.53

5=1 5/12 or 1.4166667

6=239.13

hope this helps

3 0
3 years ago
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Please help 20 points
frozen [14]

Answer: y = 42°

Step-by-step explanation: You can find the second angle in that triangle because it is next to another 90° angle. Then you add 90 and 48 and subtract that number from 180 and you get 42. Hope this helps!

4 0
3 years ago
Read 2 more answers
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