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2 years ago

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The width of a rectangle is 20m less than twice its length. If the perimeter of the rectangle is more than 56m, find the minimum

dimensions of the rectangle if each dmension in meters is an integer.

1 answer:

geniusboy [140]2 years ago

3 0

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For the figure below the base area and the height has been provided. find the VOLUME

Inga [223]

Answer:

Step-by-step explanation:

To find the volume of this solid, multiply the base area by the height:

V = (25 cm²)(4 cm) = 100 cm³

4 0

2 years ago

Use the given information to find the p-value. also use a0.05 significance level and state the conclusion about the null hypothe

omeli [17]

Answer:

(a) C: 0.0465 ; reject the null hypothesis

(b) B: 0.3409 ; fail to reject the null hypothesis

Step-by-step explanation:

Firstly, the decision rule based on P-value to reject the null hypothesis or fail to reject the null hypothesis is given by;

If the P-value of test statistics is less than the level of significance, then we have sufficient evidence to reject our null hypothesis.

If the P-value of test statistics is more than the level of significance, then we have insufficient evidence to reject our null hypothesis due to which we fail to reject our null hypothesis.

(a) Now, we are given alternate hypothesis, $H_1$ : p < 3/5

Also, z test statistics is -1.68. Since this is a left tailed test, so P-value is given by;

P-value = P(Z < -1.68) = 1 - P(Z $\leq$ 1.68)

= 1 - 0.95352 = <u>0.0465</u>

Since, the P-value of test statistics is less than the level of significance as 0.0465 < 0.05, so we have sufficient evidence to reject our null hypothesis due to which <u>we reject the null hypothesis.</u>

(b) Now, we are given alternate hypothesis, $H_1$ : p > 0.383

Also, z test statistics is 0.41. Since this is a right tailed test, so P-value is given by;

P-value = P(Z > 0.41) = 1 - P(Z $\leq$ 0.41)

= 1 - 0.6591 = <u>0.3409</u>

Since, the P-value of test statistics is more than the level of significance as 0.3409 > 0.05, so we have insufficient evidence to reject our null hypothesis due to which <u>we fail to reject the null hypothesis.</u>

5 0

2 years ago

Water is leaking out of an inverted conical tank at a rate of 6800 cubic centimeters per min at the same time that water is bein

ivanzaharov [21]

Answer:

1508527.582 cm³/min

Step-by-step explanation:

The net rate of flow dV/dt = flow rate in - flow rate out

Let flow rate in = k. Since flow rate out = 6800 cm³/min,

dV/dt = k - 6800

Now, the volume of a cone V = πr²h/3 where r = radius of cone and h = height of cone

dV/dt = d(πr²h/3)/dt = (πr²dh/dt)/3 + 2πrhdr/dt (since dr/dt is not given we assume it is zero)

So, dV/dt = (πr²dh/dt)/3

Let h = height of tank = 12 m, r = radius of tank = diameter/2 = 3/2 = 1.5 m, h' = height when water level is rising at a rate of 21 cm/min = 3.5 m and r' = radius when water level is rising at a rate of 21 cm/min

Now, by similar triangles, h/r = h'/r'

r' = h'r/h = 3.5 m × 1.5 m/12 m = 5.25 m²/12 m = 2.625 m = 262.5 cm

Since the rate at which the water level is rising is dh/dt = 21 cm/min, and the radius at that point is r' = 262.5 cm.

The net rate of increase of water is dV/dt = (πr'²dh/dt)/3

dV/dt = (π(262.5 cm)² × 21 cm/min)/3

dV/dt = (π(68906.25 cm²) × 21 cm/min)/3

dV/dt = 1447031.25π/3 cm³

dV/dt = 4545982.745/3 cm³

dV/dt = 1515327.582 cm³/min

Since dV/dt = k - 6800 cm³/min

k = dV/dt - 6800 cm³/min

k = 1515327.582 cm³/min - 6800 cm³/min

k = 1508527.582 cm³/min

So, the rate at which water is pumped in is 1508527.582 cm³/min

5 0

3 years ago

Todea what did one pencil say to another mathematics algebra order of operations answer

Serjik [45]

Part 1

$6- \frac{18-3^2}{4+(2-3)} =6- \frac{18-9}{4+(-1)} \\ \\ =6- \frac{9}{4-1} =6- \frac{9}{3} =6-3 \\ \\ =3$

Part 2

$7-8\div4(3^2-1)=7-8\div4(9-1) \\ \\ =7-8\div4(8)=7-8\div32=7- \frac{1}{2} \\ \\ =6 \frac{1}{2} = \frac{13}{2}$

Part 3

$2-2(2-5)^2+3^2=2-2(-3)^2+9 \\ \\ =2-2(9)+9=2-18+9=-7$

Part 4

$(5^2-9\cdot3)^2-11=(25-27)^2-11 \\ \\ =(-2)^2-11=4-11=-7$

Part 5

$[(2-3)^2-5]\cdot4=[(-1)^2-5]\cdot4 \\ \\ =(1-5)\cdot4=-4\cdot4=-16$

Part 6

$12-11\cdot2+16\div8=12-22+2 \\ \\ =-8$

Part 7

$-5+1\cdot3-(7-2^3)=-5+3-(7-8) \\ \\ =-2-(-1)=-2+1=-1$

Part 8

$8+2\cdot3-14\div7=8+6-2 \\ \\ =12$

Part 9

$(8-2)^2+\frac{1}{4}[4-3(6-10)]=6^2+\frac{1}{4}[4-3(-4)] \\ \\ =36+\frac{1}{4}[4-(-12)]=36+\frac{1}{4}(4+12)=36+\frac{1}{4}(16) \\ \\ =36+4=40$

Part 10

$7+(2-3^2)\cdot5=7+(2-9)\cdot5 \\ \\ =7+(-7)\cdot5=7+(-35)=7-35 \\ \\ =-28$

Part 11

$3-2[2^3+(-1)^3]=3-2[8+(-1)] \\ \\ =3-2(8-1)=3-2(7)=3-14 \\ \\ -11$

Part 12

$18+6\div3(2^2+5)=18+6\div3(4+5) \\ \\ =18+6\div3(9)=18+6\div27=18+ \frac{2}{9} \\ \\ =18 \frac{2}{9}$

8 0

3 years ago

What is the gradient of the graph shown

gavmur [86]

Answer:

Use the formula change in y divided by change in x to get your answer

8 0

2 years ago