What is 1,289,845 rounded to the nearest ten thousand?

What is the answer to 7^3 x 7^-5? I need it for an important test that I can’t fail. Please help.

Vilka [71]

The answer fir sure is 1/49

4 0

2 years ago

What are the solutions to the quadratic equation below? x^2+34x-72=0

Annette [7]

Answer:

( x + 36 ) ( x - 2 ) = 0

Step-by-step explanation:

x^2 + 36x - 2x -72 = 0

x ( x + 36 ) - 2 ( x + 36 ) = 0

( x + 36 ) ( x - 2 ) = 0

7 0

3 years ago

A house wife forgot her bank'Atm pin'which is a four digit number,but luckily she remembered some hints on how to recall this 'p

Liono4ka [1.6K]

Answer:

4829

Step-by-step explanation:

If the first digit is half of the second, that means the second is 2 times the first

So let's solve this algebraically...

The first digit could be called a

The second digit would have to be 2 times a (2a)

We can call the 3rd digit b

Since the sum of the second and third digit is 10, the equation would be this..

2x + b = 10, which is the same as b = 10 - 2x

The fourth digit is this equation

2x + 1

So the four digits are...

x, 2x, 10-2x, 2x+1

They are all equal to 23, which means...

x + 2x + 10-2x + 2x +1 = 23

After simplifying...

3x + 11 = 23

3x = 23 - 11

Which means...

3x = 12, x = 4

After you plug all that back into the prior equation you should get

first digit = 4 second digit = 8 third digit = 2 fourth digit = 9

Thank me later :)

(or just mark brainliest)

6 0

3 years ago

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean. See Attached Excel for Da

alexgriva [62]

Answer:

The 99% confidence interval for the mean germination time is (12.3, 19.3).

Step-by-step explanation:

The question is incomplete:

Recorded here are the germination times (in days) for ten randomly chosen seeds of a new type of bean: 18, 12, 20, 17, 14, 15, 13, 11, 21, 17. Assume that the population germination time is normally distributed. Find the 99% confidence interval for the mean germination time.

We start calculating the sample mean M and standard deviation s:

$M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{10}(18+12+20+17+14+15+13+11+21+17)\\\\\\M=\dfrac{158}{10}\\\\\\M=15.8\\\\\\$

$s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{9}((18-15.8)^2+(12-15.8)^2+(20-15.8)^2+. . . +(17-15.8)^2)}\\\\\\s=\sqrt{\dfrac{101.6}{9}}\\\\\\s=\sqrt{11.3}=3.4\\\\\\$

We have to calculate a 99% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=15.8.

The sample size is N=10.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

$s_M=\dfrac{s}{\sqrt{N}}=\dfrac{3.4}{\sqrt{10}}=\dfrac{3.4}{3.162}=1.075$

The degrees of freedom for this sample size are:

df=n-1=10-1=9

The t-value for a 99% confidence interval and 9 degrees of freedom is t=3.25.

The margin of error (MOE) can be calculated as:

$MOE=t\cdot s_M=3.25 \cdot 1.075=3.49$

Then, the lower and upper bounds of the confidence interval are:

$LL=M-t \cdot s_M = 15.8-3.49=12.3\\\\UL=M+t \cdot s_M = 15.8+3.49=19.3$

The 99% confidence interval for the mean germination time is (12.3, 19.3).

8 0

3 years ago

Can Someone please help me solve this

Goryan [66]

Expression: 3(x - 5) = 21

= 3x - 15 = 21

= 3x = 21 + 15

= 3x = 36

= x = 36/3

= x = 12

In short, Your Answer would be: 12

Hope this helps!

6 0

3 years ago