YO VIC If y=56 when x=49 find y when x=81

Separate 60 into two parts so that 3 times the smaller added to 6 more than 6 times the smaller = 60.

spin [16.1K]

Our number is going to be 6.

5 0

3 years ago

Read 2 more answers

2.- Supóngase que los diámetros de los tornillos fabricados por una compañía están distribuidos normalmente con una media de 0.2

Sergio [31]

Answer:

7.30167%

Step-by-step explanation:

Usando la fórmula de puntuación z

z = (x-μ) / σ, donde x es la puntuación bruta, μ es la media de la población y σ es la desviación estándar de la población

Para x <0.20 pulgadas

z = 0.20 - 0.25 / 0.02

z = -2.5

Valor de probabilidad de Z-Table:

P (x <0.20) = 0.0062097

Para x> 0.28 pulgadas

z = 0.28 - 0.20 / 0.02

z = 1.5

Valor de probabilidad de Z-Table:

P (x <0.28) = 0.93319

P (x> 0.28) = 1 - P (x <0.28) = 0.066807

La probabilidad de que se produzcan tornillos defectuosos cuando el tornillo se considera defectuoso si su diámetro es inferior a 0.20 pulgadas o superior a 0.28 pulgadas es

P (x <0.20) + P (x> 0.28)

= 0.0062097 + 0.066807

= 0.0730167

Conversión a porcentaje

= 0.0730167 × 100

= 7.30167%

El porcentaje de tornillos defectuosos producidos es

7.30167%

3 0

3 years ago

Y= -3x+5 please solve

pychu [463]

X can be from -1.6 to -1.7 or x can be from -3.3 to 3.3

6 0

3 years ago

Read 2 more answers

It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are rando

notka56 [123]

Answer:

It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

Step by Step Solution:

The given data are;

Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

Using the functions of Microsoft Excel, we get;

The mean hourly rate for fast-food Chain A, $\overline x_1$ = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, $\overline x_2$ = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

The null hypothesis, H₀: $\overline x_1$ = $\overline x_2$

The alternative hypothesis, Hₐ: $\overline x_1$ ≠ $\overline x_2$

The pooled variance, $S_p^2$ , is given as follows;

$S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}$

Therefore, we have;

$S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682$

The test statistic is given as follows;

$t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}$

Therefore, we have;

$t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176$

The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

Therefore, given that the absolute value of the test statistic is less than the critical 't', we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

3 0

3 years ago

An episode of a television show is 60 minutes long when it originally airs with commercials. On a DVD without commercials, the e

ryzh [129]

60-41 1/2=18 1/2 18 1/2 divided by 1/2=37/2*2 (cross cancel 2s)=37

8 0

3 years ago