Whats the measurment for H?

-26 times a number minus 22 is equal to 90 less than the number.

IgorLugansk [536]

Answer:

$x = \frac{68}{27}$

Step-by-step explanation:

Let x be the number

Write it out: -26x - 22 = x - 90
Combine like terms: -27x + 68 = 0
Subtract 68 from each side, so it now looks like this: -27x = -68
Divide each side by -27 to cancel out the -27 next to x. It should now look like this: $x = \frac{68}{27}$

I hope this helps!

6 0

2 years ago

Pls help this is an ixl question and it really is hard

valentina_108 [34]

Answer:

It would be the first one... 10:25

Step-by-step explanation:

10/4=2.5

2.5*10=25

2.5*4=10

7 0

2 years ago

The population of a town in 2000 was 430. The population is increasing at a rate of 0.9% every year. What will be the projected

never [62]

The formula is
P=Ae^rt
P ?
A 430
R 0.9/100=0.009
T time 2010-2000=10 years
E constant
P=430×e^(0.009×10)
p=470

5 0

2 years ago

1. A bag contains rubber bands with lengths that are normally distributed with a mean of 6 cm of length, and a standard deviatio

romanna [79]

Roughly 1.7 percent of the bands are shorter than 3cm. We calculate the z score of the data point in standard distribution. By definition of z score, we use score minus mean divided by standard deviation. z=(3-6)/1.5=-2. A z score of -2 corresponds to approximately 1.7%, in other words, roughly 1.7 percent of data is less than 3cm.

4 0

3 years ago

1. Approximate the given quantity using a Taylor polynomial with n3.

Jet001 [13]

Answer:

See the explanation for the answer.

Step-by-step explanation:

Given function:

$f(x) = x^{1/4}$

The n-th order Taylor polynomial for function f with its center at a is:

$p_{n}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(n)}a}{n!} (x-a)^{n}$

As n = 3 So,

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{3!} (x-a)^{3}$

$p_{3}(x) = f(a) + f'(a) (x-a)+\frac{f''(a)}{2!} (x-a)^{2} +...+\frac{f^{(3)}a}{6} (x-a)^{3}$

$p_{3}(x) = a^{1/4} + \frac{1}{4a^{ 3/4} } (x-a)+ (\frac{1}{2})(-\frac{3}{16a^{7/4} } ) (x-a)^{2} + (\frac{1}{6})(\frac{21}{64a^{11/4} } ) (x-a)^{3}$

$p_{3}(x) = 81^{1/4} + \frac{1}{4(81)^{ 3/4} } (x-81)+ (\frac{1}{2})(-\frac{3}{16(81)^{7/4} } ) (x-81)^{2} + (\frac{1}{6})(\frac{21}{64(81)^{11/4} } ) (x-81)^{3}$

$p_{3} (x)$ = 3 + 0.0092592593 (x - 81) + 1/2 ( - 0.000085733882) (x - 81)² + 1/6

(0.0000018522752) (x-81)³

$p_{3} (x)$ = 0.0092592593 x - 0.000042866941 (x - 81)² + 0.00000030871254

(x-81)³ + 2.25

Hence approximation at given quantity i.e.

x = 94

Putting x = 94

$p_{3} (94)$ = 0.0092592593 (94) - 0.000042866941 (94 - 81)² +

0.00000030871254 (94-81)³ + 2.25

= 0.87037 03742 - 0.000042866941 (13)² + 0.00000030871254(13)³ +

2.25

= 0.87037 03742 - 0.000042866941 (169) +

0.00000030871254(2197) + 2.25

= 0.87037 03742 - 0.007244513029 + 0.0006782414503 + 2.25

$p_{3} (94)$ = 3.113804102621

Compute the absolute error in the approximation assuming the exact value is given by a calculator.

Compute $\sqrt[4]{94}$ as $94^{1/4}$ using calculator

Exact value:

$E_{a}$ (94) = 3.113737258478

Compute absolute error:

Err = | 3.113804102621 - 3.113737258478 |

Err (94) = 0.000066844143

If you round off the values then you get error as:

|3.11380 - 3.113737| = 0.000063

Err (94) = 0.000063

If you round off the values up to 4 decimal places then you get error as:

|3.1138 - 3.1137| = 0.0001

Err (94) = 0.0001

4 0

3 years ago