Help me! Please :) waaaaaaa

I had $80.45 in my bank account on Monday. I deposited $20.50 on Tuesday, Wednesday, and Thursday. I then withdrew $37.25 on Fri

brilliants [131]

Answer:

$104.70

Step-by-step explanation:

The equation would be set up like this: 80.45 + 20.50(3) - 37.25. You started off with $80.45 in your bank account and deposited, or added, $20.50 every day on Tuesday, Wednesday and Thursday. That would mean you added $20.50 three times. Adding $80.45 + $20.50 + $20.50 + $20.50, simplified to $80.45 + $20.50(3) would get you $141.95 in total. Then, on Friday, you withdraw $37.25, getting the equation $141.95 - $37.25, leaving $104.70 for the weekend.

8 0

2 years ago

Read 2 more answers

The competitive advantage of small American factories such as Tolerance Contract Manufacturing lies in their ability to produce

Tom [10]

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Please find the complete question in the attached file.

$H_0: \sigma^{2} \leq 0.0008\\\\H_a: \sigma^{2} > 0.0008$

The testing states value is:

$\to x^2=\frac{(n-1)s^2}{\sigma^2}=32.6250$

therefor the $\rho - \ value = 0.2931$

Through out the above equation its values Doesn't rejects the H_0 value, and its sample value doesn't support the claim that although the configuration of its dependent variable has been infringed.

5 0

1 year ago

Do these numbers 19.657 < 19.67

nexus9112 [7]

I dont understand your question.......

5 0

2 years ago

Solve c =(a² + 3b)/4 for b. Bruv take all my points I literally blow at math :'(

Sati [7]

Answer:

$b=\frac{4c-a^{2} }{3}$

Step-by-step explanation:

First: multiply both sides by 4. 4 times c is 4c and the other sides cancels out as you are doing the inverse operation of division

Then, you have $4c=a^{2} +3b$

Subtract both sides by a squared

DO NOT TAKE THE SQUARE ROOT! This is because it is a squared plus 3b so you have to do the inverse of addition

From that you get $4c-a^{2} =3b$

Finally, divide both sides by 3.

You get $b=\frac{4c-a^{2} }{3}$

4 0

2 years ago

One number is 1 more than 7 times the other number. If their sum is 73 more than

kvv77 [185]

<h3>Answer:</h3>

$\boxed{\boxed{\pink{\bf The \ two \ numbers \ are \ 12 \& \ 85 . }}}$

<h3>Step-by-step explanation:</h3>

Given that , One number is 1 more than 7 times the other number. If their sum is 73 more than

twice the smaller number. And we need to find the numbers. So let us take the

First Number be x .
Second number be 7x+1 .

$\underline{\red{\boldsymbol{ According \ to \ the \ Question :- }}}$

$\bf\implies 2( Small \ no.) + 73 = Bigger \ no. + Smaller \ no.\\\\\bf\implies 2x + 73 = x + 7x + 1\\\\\bf\implies 2x + 73 = 8x + 1 \\\\\bf\implies 8x - 2x = 73-1 \\\\\bf\implies 6x = 72 \\\\\bf\implies x =\dfrac{72}{6} \\\\\bf\implies \boxed{\red{\bf x = 12 }}$

Hence the smaller number is 12 and the bigger number will be :-

$\bf \implies Number_{bigger} = 7x + 1 \\\\\bf\implies Number_{bigger} =7(12)+1\\\\\bf\implies Number_{bigger} = 84 + 1 \\\\\bf\implies \boxed{\red{\bf Number_{bigger}=85 }}$

<h3>Hence the bigger number is 85 and the smaller number is 12. </h3>

5 0

2 years ago

Read 2 more answers