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postnew [5]
4 years ago
14

What is the answer if I multipy 14x 14.50

Mathematics
2 answers:
Fudgin [204]4 years ago
7 0

Answer:

Step-by-step explanation:

203

strojnjashka [21]4 years ago
5 0

Answer:

203

Step-by-step explanation:

14×14.50=203 I calculated it

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Please help guys!!! due soon
MaRussiya [10]

Answer:

6

Step-by-step explanation:

37/37+5=1+5=6

3 0
3 years ago
Which expression is equivalent to x6 - 16x^2?
Mrac [35]

Answer:

x^2(x - 2)(x + 2)(x^2 + 4)

Step-by-step explanation:

x^6-16x^2

x^2\left(x^4-16\right)

x^2\left(x^2+4\right)\left(x+2\right)\left(x-2\right)

\boxed{\bold{HOPE \:IT \:HELPS!\: :)}}

6 0
3 years ago
"What is the value of-3|15-s|+2^3 when s-=3"?
Serggg [28]
You probably made a typing error in writing s = -3 and wrote s - = 3 instead.

The given expression is:

-3|15-s|+ 2^{3}  \\  \\ 
-3|15-s|+8


Substituting the given value of s in the previous equation, we get:

-3|15-(-3)|+8 \\  \\ 
=-3|15+3|+8 \\  \\ 
=-3|18|+8 \\  \\ 
=-54+8 = -46

Thus the value of given expression becomes -46 for s = -3
3 0
3 years ago
Help quick! The table below shows the water temperature at certain depths of a lake. Using an logarithmic model, write an equati
Norma-Jean [14]

Answer:

we need the graph lol

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Sat scores. sat scores of students at an ivy league college are distributed with a standard deviation of 250 points. two statist
erik [133]

The confidence interval for confidence level of 1-\alpha is

\left(\overline x-Z_{\alpha/2}\dfrac\sigma{\sqrt n},\overline x+Z_{\alpha/2}\dfrac\sigma{\sqrt n}\right)

where \overline x is the sample mean, Z_{\alpha/2} is the critical value for the given confidence level, \sigma is the standard deviation of the population, and n is the sample size. The margin of error is the Z_{\alpha/2}\dfrac\sigma{\sqrt n} term.

a) For a confidence level of 1-\alpha=0.90, we have Z_{\alpha/2}=Z_{0.05}\approx1.64. So in order to have a margin of error of at most 25 points, we have

1.64\dfrac{250}{\sqrt n}=25\implies n\approx268.96

so Raina should collect a sample of at least 269 students.

b) A confidence interval with a higher confidence level would more closely approximate and reflect the population, so it stands to reason that Luke should collect a larger sample than Raina to meet his 99% confidence spec.

c) For a confidence level of 1-\alpha=0.99, we have Z_{\alpha/2}=Z_{0.005}\approx2.58. Then the margin of error would at most satisfy

2.58\dfrac{250}{\sqrt n}=25\implies n\approx665.64

so that Luke should collect a sample of at least 666 students.

4 0
3 years ago
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