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3 years ago

A number divided by five is at least four

Mathematics

2 answers:

Mrrafil [7]3 years ago

7 0

Answer: 20

Step-by-step explanation: because you can divide 20 divided by 5 equals 4

tresset_1 [31]3 years ago

7 0

Answer:

x/5≥4

x≥20

Step-by-step explanation:

Multiply both sides by 5.

5x/5≥4*5

Simplify

x≥20

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Help me out please!!!!

KonstantinChe [14]

Answer:

Step-by-step explanation:

$- \frac{7}{ - f} \: is \: equal \: to \: \frac{7}{f}$

$so \: the \: answer \: is \: none \: of \: the \: above$

7 0

3 years ago

Read 2 more answers

Hi please help me!! thanks

Artist 52 [7]

Answer:

3.2 x 10^13

Step-by-step explanation:

(8x10^4)(4x10^8)

reorder the terms using the commutative property

(8x4)x(10^4x10^8)

multiply

32x (10^4 x 10^8)

= 32 x 10^4+8

= 32 x 10^12

now convert to scientific notation

=3.2 x 10^13

4 0

2 years ago

En una tienda se anuncia la rebaja de 20% en todos sus artículos . Rafael elije una camisa y al llegar a la caja le informan que

dimulka [17.4K]

Responder:

$ 750

Explicación paso a paso:

Dado que :

Descuento en todos los artículos = 20%

Descuento adicional dado = 15%

Monto final pagado por la camisa = $ 510

Precio de la camiseta sin descuento:

Sea el precio inicial de la camisa = x

Primer descuento obtenido = 20%

Precio pagado después de un descuento adicional del 15%

Primer descuento:

(1 - 0,20) * x = 0,8x

Segundo descuento:

(1 - 0,15) * 0,8x = 510

0,85 * 0,8x = 510

0,68x = 510

x = 510 / 0,68

x = $ 750

7 0

2 years ago

Points J, K and L are collinear with J between L and K. If KJ = 2x-3, LJ= 4x-8, solve for x:

notsponge [240]

Your question is incomplete, here is the complete form.

Points J, K and L are collinear with J between L and K. If KJ = 2x - 3, LK = 9x + 7 and LJ = 4x - 8, solve for x:

Answer:

The value of x is -6 ⇒ B

Step-by-step explanation:

∵ J, K, and L are collinear

→ That means they form a straight segment

∵ J is between K and L

→ That means J divides LK into two segments KJ and LJ

∴ LK = KJ + LJ

∵ LK = 9x + 7

∵ KJ = 2x - 3

∵ LJ = 4x - 8

→ Substitute them in the equation above

∴ 9x + 7 = (2x - 3) + (4x - 8)

→ Add the like terms in the right side

∵ 9x + 7 = (2x + 4x) + (-3 - 8)

∴ 9x + 7 = 6x + -11

∴ 9x + 7 = 6x - 11

→ Subtract 7 from both sides

∵ 9x + 7 - 7 = 6x - 11 - 7

∴ 9x = 6x - 18

→ Subtract 6x from both sides

∵ 9x - 6x = 6x - 6x - 18

∴ 3x = -18

→ Divide both sides by 3

∵ $\frac{3x}{3}=\frac{-18}{3}$

∴ x = -6

∴ The value of x is -6

8 0

2 years ago

A random sample of n observations is selected from a normal population to test the null hypothesis that muμequals=10. Specify th

brilliants [131]

Answer:

Step-by-step explanation:

a) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 13

Since α = 0.01, the critical value is determined from the t distribution table. Recall that this is a two tailed test. Therefore, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.01/2 = 1 - 0.005 = 0.995

The critical value is 3.012

The rejection region is area > 3.012

b) Ha: μ > 10

This is a right tailed test

n = 23

α = 0.1

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.319

The rejection region is area > 1.319

c) Ha: μ > 10

This is a right tailed test

n = 99

α = 0.05

We would reject the null hypothesis if the test statistic is greater than the table value of 1 - α

1 - α = 1 - 0.05 = 0.95

The critical value is 1.66

The rejection region is area > 1.66

d) Ha: μ < 10

This is a left tailed test

n = 11

α = 0.1

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.1 = 0.9

The critical value is 1.363

The rejection region is area < 1.363

e) H0: μ = 10

Ha: μ ≠ 10

This is a two tailed test

n = 20

Since α = 0.05, we would find the critical value corresponding to 1 - α/2 and reject the null hypothesis if the absolute value of the test statistic is greater than the value of t 1 - α/2 from the table.

1 - α/2 = 1 - 0.05/2 = 1 - 0.025 = 0.975

The critical value is 2.086

The rejection region is area > 2.086

f) Ha: μ < 10

This is a left tailed test

n = 77

α = 0.01

We would reject the null hypothesis if the test statistic is lesser than the table value of 1 - α

1 - α = 1 - 0.01 = 0.99

The critical value is 2.376

The rejection region is area < 2.376

7 0

2 years ago