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maxonik [38]
3 years ago
14

Lily has 3500 text per month on her cell phone plan up to how many more task can she spend if she already spent 2612 text

Mathematics
2 answers:
bazaltina [42]3 years ago
8 0

Answer:

888 more Text

Step-by-step explanation:

maria [59]3 years ago
4 0

Answer: She has up to 888 texts left

Step-by-step explanation:

3500-2612=888

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2x-ay=3<br><br>8x-12y=b I'm supposed to find what a-b equals.
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Answer:

2

Step-by-step explanation:

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3 years ago
Solve for C a = b + c/d<br> a. c=ad-b<br> b. c= a+b/d<br> c. c=ad/b<br> d. c= ab+d
maksim [4K]

                                                         <u>a = b + c/d</u>

Subtract  'b'  from each side:      a - b  =  c/d

Multiply each side by  'd' :      <em>d (a - b) = c

</em>
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3 years ago
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What expression is equivalent to -10x + 5
In-s [12.5K]

Answer:

-5x

Step-by-step explanation:

-10+5= -5 then you put -5 with x to get -5x

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3 years ago
Which of the following is 16 60 in simplest form
Dafna1 [17]
There are no options given, but I will give you the exact answer of 16/60 simplified.

\frac{16}{60} \\ \\ GCF = 4 \\ \\ 16 \div 4 = 4 \\ \\ 60 \div 4 = 15 \\ \\  \frac{4}{15} \\ \\ Answer: \fbox {4/15} \ or \ \fbox {0.2667}
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The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article "Testing P
zysi [14]

Answer:

Step-by-step explanation:

Hello!

Given the variable

X: breakdown voltage of an insulating liquid under certain conditions.

a)

Graphic in attachment.

Box: The median is closer to the third quartile than the first quartile showing skewness to the left.

The black dot in the middle of the box represents the sample mean (X[bar]) as you can see it is also moved to the right side of the distribution, affected by the presence of an outlier.

Whiskers:

The upper whisker is smaller than the lower one and there is a value that can be considered an outlier.

Altogether you can say that the distribution of the data set is skewed to the left.

b)

To estimate the population mean you need the population to be at least normal. The box plot shows that the distribution is not symmetrical but, considering that the sample size is n= 48 you can apply the Central Limit Theorem and approximate the sampling distribution to normal.

X[bar]≈N(μ;σ²)

As the distribution is approximate, using the sample standard deviation (since we don't know the population value) for the distribution standard deviation is also valid ⇒ Z= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } }

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

∑X= 2626 ∑X²= 144950 n= 48

X[bar]= ∑X/n= 2626/48= 54.71

S²= 1/n[∑X²- (∑X)²/n]= 1/47[144950-(2626)²/48]= 27.36

S= 5.23

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

54.71 ± 1.96 * \frac{5.23}{\sqrt{48} }

[53.22; 56.18]kV

c)

95% CI

amplitude a=2 kV  i.e. semi amplitude d= 1 kV

The semi amplitude or margin of error for the CI can be calculated as:

d= Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

From this formula you can clear the sample size

\frac{d}{Z_{1-\alpha /2}} = \frac{S}{\sqrt{n} }

(\frac{d}{Z_{1-\alpha /2}} )*\sqrt{n} = S

\sqrt{n} = S* (\frac{Z_{1-\alpha /2}}{d} )

n = (S* (\frac{Z_{1-\alpha /2}}{d} ))^2

n= (5.23*(\frac{1.96}{1} ))^2= 105.08= 106

3 0
3 years ago
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