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Dmitriy789 [7]
3 years ago
5

What is w (6) for the function w (x)=4x+5

Mathematics
2 answers:
Sedbober [7]3 years ago
7 0
W(6)
=4x+5
= <span>4(6)+5
= 24 + 5
=29</span>
deff fn [24]3 years ago
5 0
If you solve for x the answer would be x = - 5/4 + 3w/2

If you solve for w the answer would be w = 2x/3 + 5/6

Hope this helped :)
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A transformation describes a change in location, orientation, or size of a figure. Translations, reflections, and rotations belo
marshall27 [118]

Answer: These are called rigid transformations because the structure and shape of the shape remains the same.

Step-by-step explanation: The word rigid and rigidity is often used to describe something that is tough or the strength of something. The transformations might be called a rigid transformation because the rigidity of the shape remains the same because the size is the exact same, it is the exact same shape it is just moved.

3 0
3 years ago
The local branch of the Internal Revenue Service spent an average of 21 minutes helping each of 10 people prepare their tax retu
AURORKA [14]

Answer:

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

c. t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

Step-by-step explanation:

When the standard deviations are not the same then the confidence intervals for mean differences are calculated as

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

x1`= 21        x2`= 27

n1=  10       n2= 14

s1= 5.6       s2= 4.3

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

The t∝/2 for 17 d.f = 2.11

Putting the values

(x1`-x2`)- t∝/2 √s1²/n1 + s2²/n2 < u1-u2 < (x1`-x2`)+ t∝/2 √s1²/n1 + s2²/n2

(21-27) - 2.11√5.6²/10+ 4.3²/14 < u1-u2 <(21-27)  +2.11√5.6²/10+4.3²/14

6- 2.11*2.111 < u1-u2 <  ( 6 )  +2.11*2.111

6- 4.4521 < u1-u2 <  ( 6 )  +5.294

- 1.5479 < u1-u2 <  10.4521

The confidence Interval is [- 0.7053  10.4521]

a: The  hypotheses  are

H0: μ1=μ2    against the claim Ha :μ1≠μ2

The claim is that there is a difference in the average time spent by the two services

b. The critical value for t∝/2 for 17 d.f  t > 2.508 and  t < -2.111

The degrees of freedom is calculated using

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

= 17

c. The test statistic is

t= (x1`-x2`)  /√s1²/n1 + s2²/n2

t= (21-27)  /√5.6²/10+ 4.3²/14

t= -6/2.111

t= -2.8422

d. The calculated value of t= -2.8422 is less than t < -2.11 the critical value therefore we reject H0 and conclude there is a difference between the two means.

5 0
3 years ago
Which table represents a nonlinear function?
Setler [38]

Answer:

A

Step-by-step explanation:

Nonlinear would be any function where both values are not changing by some fixed amount. In B, f(x) increases by 1 for every time x increases by 1 so that's linear. For C, f(x) increases by 2 for every time x increases by 1, so that's linear. For D, f(x) increases by 3 for every time x increases by 1 so that's also linear. For A, f(x) does not always change by the same amount, so it's nonlinear.

6 0
3 years ago
What is 7/10+1/4 in simplest form
Stells [14]

Answer: 19/20

Step-by-step explanation:

7/10 turns into 14/20

1/4 turns into 5/20

14/20 + 5/20

14 + 5 over 20

= 19/20

19/20 is in simplest form.

3 0
4 years ago
Read 2 more answers
At what rate per annum ci will RS 2000 amount to RS 2315.35 in 3 years
Mashutka [201]

<u>ANSWER: </u>

Rate per annum at which CI will amount from RS 2000 to RS 2315.35 in 3 years is 5%

<u>SOLUTION: </u>

Given,  

P = RS 2000

C.I = RS 2315.35

T = 3 years

We need to find the rate per annum. i.e. R = ?

We know that,  

When interest is compound Annually:

Amount $=\mathrm{P}\left(1+\frac{\mathrm{R}}{100}\right)^{n}$

Where p = principal amount

r = rate of interest

n = number of years

$2315.35=2000 \times\left(1+\frac{R}{100}\right)^{3}$

$\left(1+\frac{R}{100}\right)^{3}=\frac{2315.35}{2000}$

$\left(1+\frac{R}{100}\right)^{3}=1.157$

$1+\frac{R}{100}=\sqrt[3]{1.157}$

$1+\frac{R}{100}=1.0500$

$\frac{R}{100}=1.05-1$

$\frac{R}{100}=0.05$

R = 5%

Hence, rate per annum is 5 percent.

3 0
3 years ago
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