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belka [17]
3 years ago
13

Determine the slope from the given graph below:

Mathematics
1 answer:
Marat540 [252]3 years ago
4 0

Answer:

<u>Slope = -2</u>

Additionally;

Y-intercept = -3

The whole equation: y = -2x - 3

Hope this helps!

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How many times longer is 3 1/2 to 2 1/2
Alik [6]

Answer:

1.4 times as long

Step-by-step explanation:

3 1/2 (convert to mixed number) --> 7/2

2 1/2 (convert to mixed number) --> 5/2

(7/2) / (5/2) --> 7/5 --> 1.4

3 0
2 years ago
Read 2 more answers
For the following right triangle, find the side length x.
lesantik [10]

Answer:

x= 5√7

Step-by-step explanation:

Use the Pythagorean theorem

16^2= 9^2 + x^2

x^2=175

x= √175

x= 5√7

6 0
3 years ago
Q.2 The lengths of a triangle are 7 cm, 8 cm and 10 cm correct to the nearest cm. Work out maximum possible perimeter of the tri
masha68 [24]

Step-by-step explanation:

7+8+10=25cm is the possibile perimeter of this

triangle.

8 0
2 years ago
Use the normal distribution to find a confidence interval for a proportion p given the relevant sample results. Give the best po
Brrunno [24]

Answer: (0.63, 7.07)

Step-by-step explanation:

The confidence interval for population proportion is given by :-

\hat{p}\pm z_{\alpha/2}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}

Given :\hat{p} =0.7 ;  n=110

Significance level : 1-0.90=0.1

Critical value :  z_{\alpha/2}=1.645

Now, a 90% confidence interval for population proportion will be :-

0.7\pm (1.645)\sqrt{\dfrac{0.7(1-0.7)}{110}}\\\\\approx0.7\pm0.07\\\\=(0.7-0.07,0.7+0.07)=(0.63,\ 7.07)

Hence, a 90% confidence interval for population proportion = (0.63, 7.07)

5 0
3 years ago
Does anyone know the answer?
laila [671]

Answer:

a. θ = 30°

b. μ = √3 / 15 ≈ 0.115

Step-by-step explanation:

Draw a free body diagram for each scenario (see attached figure).  The body has four forces acting on it:

  • Weight pulling down
  • Normal force perpendicular to the incline
  • Applied force parallel up the incline
  • Friction force parallel to the incline

Remember that friction opposes the direction of motion.  So when the body is sliding up, friction points down the incline.  And when the body is sliding down, friction points up the incline.

Now apply Newton's second law to each scenario, first in the normal direction, then in the parallel direction.

For sliding up, sum of the forces normal to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Sum of the forces parallel to the incline:

∑F = ma

P₁ − f − mg sin θ = 0

P₁ − Nμ − mg sin θ = 0

Substituting the expression for normal force:

P₁ − mgμ cos θ − mg sin θ = 0

Now for sliding down, sum of the forces normal to the incline:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

And sum of the forces parallel to the incline:

∑F = ma

P₂ + f − mg sin θ = 0

P₂ + Nμ − mg sin θ = 0

Substituting the expression for normal force:

P₂ + mgμ cos θ − mg sin θ = 0

We know that P₁ = 6 kg.wt, P₂ = 4 kg.wt, and mg = 10 kg.wt.

So we have two equations and two unknowns (μ and θ):

P₁ − mgμ cos θ − mg sin θ = 0

P₂ + mgμ cos θ − mg sin θ = 0

Let's start by adding the equations together:

P₁ + P₂ − 2 mg sin θ = 0

P₁ + P₂ = 2 mg sin θ

sin θ = (P₁ + P₂) / (2 mg)

Plugging in the values:

sin θ = (6 + 4) / (2 × 10)

sin θ = 1/2

θ = 30°

Now we can plug this into either equation and find μ.

P₁ − mgμ cos θ − mg sin θ = 0

6 − (10 cos 30°) μ − 10 sin 30° = 0

6 − 5√3 μ − 5 = 0

1 = 5√3 μ

μ = √3 / 15

μ ≈ 0.115

6 0
3 years ago
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