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3 years ago

H - 8 = 2; 28 tell whether the given number is a solution to the equation

Mathematics

1 answer:

murzikaleks [220]3 years ago

8 0

H-8=2. The only solution to that equation is H=10. Any other number isn't.

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In an isosceles triangle, the angle formed by its legs is called__

Alecsey [184]

Answer:

a:vertex angle

Step-by-step explanation:

it is vertex angle

7 0

3 years ago

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Identify the 7th term of the geometric sequence in which a2 = 324 and a4 = 36.

ziro4ka [17]

Answer:

a7=4/3 or a7=-4/3

Step-by-step explanation:

using the geometric sequences formula

an=ar^n-1

a2=ar

a4=ar³

when a2=324 and a4=36

324=ar...........(1)

36=ar³............(2)

from equation (1) a=324/r substitute in equation (2)

we have :

36=324/r *r³

36=324r²

r²=36/324

r²=1/9

r=±1/3

substitute when r=±1/3 in (1)

324=a(±1/3)

a=±972

so the 7th term is

when r=±1/3

we have

a7=ar^6

a7=±972(±1/3)^6

a7=972/729

a7=4/3 or a7=-4/3

3 0

2 years ago

I NEED HELP! Let me know how you got that answer I’m trying to understand and I’ll mark you brainliest :)

zlopas [31]

Answer:

A- (-1,0) B- (1,7) C- (6,5)

Step-by-step explanation:

these would be your new coordinates

6 0

4 years ago

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A population of 50 deer are introduced into a wildlife sanctuary. It is estimated that the sanctuary can sustain up to 400 deer.

Mars2501 [29]

Answer:

p1= 95, p2 = 180

Step-by-step explanation:

We have population : 50 given that population grows by 90% per year

So, After one year: new population (p1) will be

p1 = 50 + 50*0.9 = 50 + 45 = 95

Now, After two years: new population (p2) will be

p2 = 95 + 95*0.9 = 95 + 85.5 (approximately = 86) = 95 + 86 = 180

Therefore,

p1 = 95, and p2 = 180.

8 0

3 years ago

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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