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Xelga [282]
4 years ago
10

Super confused help me out?

Mathematics
1 answer:
balu736 [363]4 years ago
8 0

Answer:9

Step-by-step explanation:

90 -28+53.... It's 90 because the angles in a right triangle add up to 90 hope this helps brainliest

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Please Hurry!!
sertanlavr [38]
6/3 = 2
10^4/10^-5 = 10^9

so it's 2 x 10^9

answer
<span>A 2 × 10^9</span>
8 0
4 years ago
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If the expression below has a positive value which inequality represents all possible values of X in the expression? -3x
prisoha [69]
-3x > 0
Divide both sides by -3 remembering to "flip" the inequality as you do so, because it's negative.
6 0
4 years ago
Find the product of 30−−√ and 610−−√. Express it in standard form (i.E., ab√).
Zina [86]

we are given

\sqrt{30} *\sqrt{610}

we can radical formula

\sqrt{a} *\sqrt{b}=\sqrt{a*b}

we get

\sqrt{30} *\sqrt{610}=\sqrt{30*610}

\sqrt{30} *\sqrt{610}=\sqrt{3*61*100}

we can  also write as

\sqrt{30} *\sqrt{610}=\sqrt{100}*\sqrt{3*61}

\sqrt{30} *\sqrt{610}=10\sqrt{183}............Answer


7 0
4 years ago
Read 2 more answers
Suppose the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65
erma4kov [3.2K]

Answer:

Probability that the sample average is at most 3.00 = 0.98030

Probability that the sample average is between 2.65 and 3.00 = 0.4803

Step-by-step explanation:

We are given that the sediment density (g/cm) of a randomly selected specimen from a certain region is normally distributed with mean 2.65 and standard deviation 0.85.

Also, a random sample of 25 specimens is selected.

Let X bar = Sample average sediment density

The z score probability distribution for sample average is given by;

               Z = \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } ~ N(0,1)

where, \mu = population mean = 2.65

           \sigma  = standard deviation = 0.85

            n = sample size = 25

(a) Probability that the sample average sediment density is at most 3.00 is given by = P( X bar <= 3.00)

    P(X bar <= 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 2.06) = 0.98030

(b) Probability that sample average sediment density is between 2.65 and 3.00 is given by = P(2.65 < X bar < 3.00) = P(X bar < 3) - P(X bar <= 2.65)

P(X bar < 3) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } < \frac{3-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z < 2.06) = 0.98030

 P(X bar <= 2.65) = P( \frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } } <= \frac{2.65-2.65}{\frac{0.85}{\sqrt{25} } } ) = P(Z <= 0) = 0.5

Therefore, P(2.65 < X bar < 3)  = 0.98030 - 0.5 = 0.4803 .

                                                                             

8 0
4 years ago
The sum of the digits of a two-digit number is 12. the number formed by interchanging the digits is 54 more than the original nu
Grace [21]
The number is xy
x+y=12

10y+x=54+10x+y

so

x+y=12
minus x both sides
y=12-x
sub that for y

10(12-x)+x=54+10x+(12-x)
120-10x+x=54+10x+12-x
120-9x=66+9x
add 9x both sides
120=66+18x
minus 66 both sides
54=18x
divide both sides by 18
3=x

sub back
y=12-x
y=12-3
y=9

the number is 39
4 0
3 years ago
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