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Yakvenalex [24]

4 years ago

9

Five-digit locker combinations are assigned using the digits 1–9. In how many ways can the combinations be formed if no digit ca

n be repeated

1 answer:

Orlov [11]4 years ago

4 0

Answer:

15, 120

Step-by-step explanation:

9P4

.............................................................................................

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I will mark u as brainliest if u answer this!!

lawyer [7]

i)On z, define a∗b=a−b
here aϵz
+
and bϵz
+

i.e.,a and b are positive integers
Let a=2,b=5⇒2∗5=2−5=−3
But −3 is not a positive integer
i.e., −3∈
/
z
+

hence,∗ is not a binary operation.
ii)On Q,define a∗b=ab−1
Check commutative
∗ is commutative if,a∗b=b∗a
a∗b=ab+1;a∗b=ab+1=ab+1
Since a∗b=b∗aforalla,bϵQ
∗ is commutative.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(ab+1)∗c=(ab+1)c+1=abc+c+1
a∗(b∗c)=a∗(bc+1)=a(bc+1)+1=abc+a+1
Since (a∗b)∗c

=a∗(b∗c)
∗ is not an associative binary operation.
iii)On Q,define a∗b=
2
ab

Check commutative
∗ is commutative is a∗b=b∗a
a∗b=
2
ab

b∗a=
2
ba

=
2
ab

a∗b=b∗a∀a,bϵQ
∗ is commutativve.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=
2
(
2
ab

)∗c

=
4
abc

(a∗b)∗c=a∗(b∗c)=
2
a×
2
bc

=
4
abc

Since (a∗b)∗c=a∗(b∗c)∀a,b,cϵQ
∗ is an associative binary operation.
iv)On z
+
, define if a∗b=b∗a
a∗b=2
ab

b∗a=2
ba
=2
ab

Since a∗b=b∗a∀a,b,cϵz
+

∗ is commutative.
Check associative.
∗ is associative if $$
(a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(2
ab
)
∗
c=2
2
ab

c
a∗(b∗c)=a∗(2
ab
)=2
a2
bc

Since (a∗b)∗c

=a∗(b∗c)
∗ is not an associative binary operation.
v)On z
+
define a∗b=a
b

a∗b=a
b
,b∗a=b
a

⇒a∗b

=b∗a
∗ is not commutative.
Check associative
∗ is associative if $$
(a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(a
b
)
∗
c=(a
b
)
c

a∗(b∗c)=a∗(2
bc
)=2
a2
bc

eg:−Leta=2,b=3 and c=4
(a∗b)
∗
c=(2∗3)
∗
4=(2
3
)
∗
4=8∗4=8
4

a∗(b∗c)=2
∗
(3∗4)=2
∗
(3
4
)=2∗81=2
81

Since (a∗b)∗c

=a∗(b∗c)
∗ is not an associative binary operation.
vi)On R−{−1}, define a∗b=
b+1
a

Check commutative
∗ is commutative if a∗b=b∗a
a∗b=
b+1
a

b∗a=
a+1
b

Since a∗b

=b∗a
∗ is not commutatie.
Check associative
∗ is associative if (a∗b)∗c=a∗(b∗c)
(a∗b)∗c=(
b+1
a

)
∗
c=
c
b
a

+1

=
c(b+1)
a

a∗(b∗c)=a∗(
c+1
b

)=
c+1
b
a

=
b
a(c+1)

Since (a∗b)∗c

=a∗(b∗c)
∗ is not a associative binary operation

6 0

3 years ago

Volume of prisms and cylinders

Mumz [18]

Answer:

7)V=16257.74

8)V= 226.19

9)V=12.56

10)V=6283.18

11)r=6.631

12)h=69.05

13)V=35328.78

Step-by-step explanation:

Volume of cylinder is given by formula:

V=πr^2h

where r is radius and h is height of cylinder

7)

r=15

h=23

Putting values in formula:

V= π(15)^2(23)

=5175π

=16257.74

8)

diameter=6

r=d/2=3

height =?

Finding height by pythagoras theorem:

h^2= 10^2-6^2

=100-36

=64

h=8

Putting values in formula:

V=π(3)^2(8)

= 72π

= 226.19

9)

r= 2

h=1

Putting values in formula:

V=π(2)^2(1)

=4π

= 12.56

10)

diameter =20

r=d/2=10

h=20

Putting values in formula:

V=π(10)^2(20)

=2000π

=6283.18

Now finding the missing measures:

11)

Given V=210π

h = 15

r=?

Putting values in formula:

210π=π(r)^2(15)

r^2= 210π/15

=14π

r= 6.631

12)

V=197.82

r=3

h=?

Putting values in formula:

197.82=π(3)^2(h)

h= 197.82/9π

= 69.05

13)

r=21

h=25.5

Putting values in formula:

V= π(21)^2(25.5)

=11245.5π

=35328.78 !

8 0

4 years ago

Ft? 1(907)-952-9189

Alex_Xolod [135]

Answer:

yessir im down

8 0

3 years ago

The concentration of hexane (a common solvent) was measured in units of micrograms per liter for a simple random sample of sixte

goldenfox [79]

Answer:

Yes, it can be concluded that the mean hexane concentration is less in treated water than in unsaturated water

Step-by-step explanation:

The number of of specimen in the samples of untreated water, n₁ = 16

The sample mean, $\overline x_1$ = 228.0

The sample standard deviation, s₁ = 4.3

The number of of specimen in the samples of treated water, n₂ = 20

The sample mean, $\overline x_2$ = 224.6

The sample standard deviation, s₂ = 5.0

The level of significance = 0.10

The null hypothesis, H₀; $\overline x_1$ ≥ $\overline x_2$

The alternative hypothesis, Hₐ; $\overline x_1$ < $\overline x_2$

The degrees of freedom = 16 - 1 = 15

The test statistic, $t_{\alpha}$ = 1.341

$t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{\dfrac{s_{1}^{2} }{n_{1}}-\dfrac{s _{2}^{2}}{n_{2}}}}$

Plugging in the values, we get;

$t=\dfrac{(224.6- 228.0)}{\sqrt{\dfrac{5.0^{2}}{20} -\dfrac{4.3^{2} }{16}}} \approx -11.0675$

Given that the t-value is large, the corresponding p-value is low, therefore, we fail to reject the null hypothesis and there is considerable statistical evidence to suggest that the mean hexane concentration is less in treated than in untreated water, therefore, we have; $\overline x_1$ ≥ $\overline x_2$

6 0

3 years ago

The length of carter's driveway is 12 m 38 cm. His neighbors driveway is 4m 99cm longer . How long is his neighbors driveway.

nevsk [136]

Just add 12 m 38 cm with 4 m 99 cm.

Answer: 17 m 37 cm

3 0

3 years ago