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4 years ago

I need help very bad please. You guys can do any questions just tell me the number

Mathematics

1 answer:

Andrews [41]4 years ago

6 0

Question 10: P=28cm

if l=8 and w=6

Then (8+6)=14
2(14) so 14x2 = 28
So therefore P=28cm

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What is the justification for step 2 in the solution process?

Contact [7]

When we bring 6y from right side to left side it’s sign get changed and 8y-6y=2y.

3 0

3 years ago

1.2x+1 for x =8 •14 •10.6 •-13 •-13.9 •6

Alina [70]

Answer:

10.6

Step-by-step explanation:

1.2×8 is 9.6 then 9.6 +1 is 10

Always Multiply or Divide first and then Add or Subtract.

5 0

4 years ago

) Solve the equation 3x²-7x-6=0 by completing the square method

galina1969 [7]

Answer:

x=3

Step-by-step explanation:

$3x^{2}$ -7x-6=0

9x-7x-6=0

=9x-7x=0+6=6

2x=6

x=6/2=3

4 0

3 years ago

The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for

Morgarella [4.7K]

Answer:

95% confidence interval for the population variance = (1.42 , 2.62).

Step-by-step explanation:

We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.

So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;

P.Q. = $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, $s^{2}$ = sample variance = 1.88

$\sigma^{2}$ = population variance

n = sample of windshields = 83

So, 95% confidence interval for population variance, $\sigma^{2}$ is;

P(58.85 < $\chi^{2} __8_2$ < 108.9) = 0.95 {As the table of $\chi^{2}$ at 82 degree of freedom

gives critical values of 58.85 & 108.9}

P(58.85 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 108.9) = 0.95

P( $\frac{ 58.85}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{108.9}{(n-1)s^{2} }$ ) = 0.95

P( $\frac{ (n-1)s^{2}}{108.9 }$ < $\sigma^{2}$ < $\frac{ (n-1)s^{2}}{58.85 }$ ) = 0.95

95% confidence interval for $\sigma^{2}$ = ( $\frac{ (n-1)s^{2}}{108.9 }$ , $\frac{ (n-1)s^{2}}{58.85 }$ )

= ( $\frac{ (83-1)\times 1.88}{108.9 }$ , $\frac{ (83-1)\times 1.88}{58.85 }$ )

= (1.42 , 2.62)

Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).

8 0

3 years ago

A bag contains pennies nickels and dimes there are 50 coins in all of the coins 16% are pennies and 36%are dimes there are 6 mor

QveST [7]

Answer:

The bag contains $2.39

Explanation 40% of 50 coins is 20 dimes which equals 2 dollars

8 percent is 4 pennies

2.00+ .04 is 2.04

Plus 7 nickels because there are 3 more nickels than pennies 7*5= 35

2.04+.35= 2.39

8 0

3 years ago