Help me with 1 and 2

People tend to speak more quietly in restaurants than they do when they are having an ordinary conversation. Restaurant conversa

In-s [12.5K]

Conversational speech is 65 dB as a known fact.

Using math, dB = 10 log[10] ratio
10x => 10(1) = +10 dB
100x => 10(2) = +20 dB
base is 45 dB

7 0

3 years ago

A company placed $500,000 in three different investment plans.

Paha777 [63]

Answer:

B. $200,000

Step-by-step explanation:

Let x be the amount placed at 5.5%

3x is the amount placed at 5%

This leaves the rest for the 4.5 % 500000- x - 3x = 500000-4x

Assuming simple interest

I = PRT where I is interest, P is principle, rate is rate and t is time

Interest for the 5.5% for 1 year is

I = x * .055 * 1 = .055x

Interest for the 5.% for 1 year is

I = 3x * .05 * 1 = .15x

Interest for the 4.5% for 1 year is

I = 500000-4x * .045 * 1 = 22500 - 0.18 x

Add the interest together to get the income for 1 year

.055x+ .15x+22500 - 0.18 x=25,000

Combine like terms

0.025x+22500=25000

Subtract 22500

.025x =2500

Divide by .025

x =100000

short term notes is 100000

3*x = 300000 = government bonds

The rest is utility bonds 500000- 100000-300000 = 100000

We want how much more in government bond than utility bond

300000 -100000 = 200000

8 0

3 years ago

Read 2 more answers

Find a polynomial of the form

Sloan [31]

Answer:

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$

Step-by-step explanation:

We are given that

$f(x)=ax^3+bx^2+cx+d$

f(0)=-3,f(1)=2,f(3)=5 and f(4)=0

We have to find the polynomial

Substitute the value x=0 then ,we get

f(0)=d=-3

Substitute x=1 then we get

$a+b+c-3=2$

$a+b+c=2+3=5$

$a+b+c=5$ (equation I)

Substitute x=3 then we get

$a(3)^3+b(3)^2+c(3)-3=5$

$27a+9b+3c=5+3=8$

$27a+9b+3c=8$ (Equation II)

Substitute x=4 then we get

$a(4)^3+b(4)^2+c(4)-3=0$

$64a+16b+4c=3$ (Equation III)

Equation I multiply by 3 then subtract from equation II

$24a+6b=-7$ (Equation IV)

Equation II multiply by 4 and equation III multiply by 3 and subtract equation II from III

$84a+12b=-23$ (Equation V)

Equation IV multiply by 2 and then subtract from equation V

$36a=-9$

$a=-\frac{9}{36}$

$a=-\frac{1}{4}$

Substitute the value of a in equation IV then we get

$24(-\frac{1}{4})+6b=-7$

$-6+6b=-7$

$6b=-7+6=-1$

$b=-\frac{1}{6}$

Substitute the value of b in equation I then we get

$-\frac{1}{4}-\frac{1}{6}+ c=5$

$-\frac{5}{12}+c=5$

$c=5+\frac{5}{12}=\frac{65}{12}$

Substitute the values then we get

$f(x)=-\frac{1}{4}x^3-\frac{1}{6}x^2+\frac{65}{12}x-3$

5 0

4 years ago

9.

artcher [175]

Answer:

C(-1, -4)

Step-by-step explanation:

see the point reflected over the line through the vertex point

8 0

3 years ago

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Point C is located at (1, 2) and point D is located at (-4, -2). Find the X value for the point that is 1/4 the distance from po

kolbaska11 [484]

D (-4; -2)
C (1; 2)
M (x; y)
DM+MC=CD
CM:MD=1:3
M: x= x(C)-(x(C) - x(D))/4=1-(1-(-4))/4=1 - (6/4)=1 - 1,5= - 0,5
M: y=y(C)-(y(C) - y(D))/4=2-(2-(-2))/4=2 - (4/4)=2 - 1=1
M(x; y)=M( -0,5; 1)

8 0

4 years ago