Mario buys a package of 4 batteries for under $2.52. what is the price range for each battery?

Click and drag like terms onto each other to simplify fully <br> -2x+1+7x-3y

erma4kov [3.2K]

After it is fully simplified, it will be 5x+1-3y

7 0

3 years ago

What is the solution to the inequality

Dvinal [7]

Answer:

p > 5 and p <-8

Step-by-step explanation:

To solve this, you first need to isolate p.

First add 6 on both sides of the equation:

$-6 + |2p+3| >7\\\\(+6) -6 + |2p+3| >7 +6\\\\2p + 3 > 13$

Then subtract 3 from both sides of the equation.

$2p+3-3>13-3\\\\2p > 10\\$

The divide both sides by 2.

$\dfrac{2p}{2}>\dfrac{10}{2}\\\\p>5$

Another solution is possible because of the absolute value.

If $|2p+3|>13$

Then $|2p+3|$

<em>Thus we can solve the second solution:</em>

$|2p+3|$

$2p+3$

Isolate P again by subtracting both sides by 3

$2p+3-3$

$2p$

Then divide both sides by 2:

$\dfrac{2p}{2}$

$p$

3 0

3 years ago

Read 2 more answers

Write an equation to find the amount of interest $450,000 earns in 1 year at a simple interest rate of 3.5%.

EleoNora [17]

Answer:

450000x.035

$15,750

..........

4 0

2 years ago

Question 12 The GPAs of all students enrolled at a large university have an approximately normal distribution with a mean of and

andreev551 [17]

Complete question :

The GPAs of all students enrolled at a large university have an approximately normal distribution with a mean of 3.02 and a standard deviation of .29.Find the probability that the mean GPA of a random sample of 20 students selected from this university is 3.10 or higher.

Answer:

0.10868

Step-by-step explanation:

Given that :

Mean (m) = 3.02

Standard deviation (s) = 0.29

Sample size (n) = 20

Probability of 3.10 GPA or higher

P(x ≥ 3.10)

Applying the relation to obtain the standardized score (Z) :

Z = (x - m) / s /√n

Z = (3.10 - 3.02) / 0.29 / √20

Z = 0.08 / 0.0648459

Z = 1.2336940

p(Z ≥ 1.2336) = 0.10868 ( Z probability calculator)

8 0

2 years ago

g 1.32 Two points on a sphere of radius 3 are given as P1(3,0,30) and P2(3,45,45): (a) Find the position vectors of P1 and P2. (

Rama09 [41]

Answer:

a) P.V of is OP₁ = [ 1.5i + 0j + 2.6k ], P.V of is OP₂ = [ 1.5i + 1.5j + 2.12k ]

b) Vector connecting P₁ to P₂ is [ 0i + 1.5j + 0.48k ]

c) cylindrical coordinates are (1.5, π/2, 0.48)

Step-by-step explanation:

Given that;

r = 3

P₁ ( 3, 0°, 30° ), P₂ ( 3, 45°, 45° )

a)

P.V of P₁

x = rcos∅sin∅ = 3(cos0°) ( sin30°) = (3 × 1 × 0.5) = 1.5

y = rsin∅sin∅ = 3(sin0°) (sin30°) = (3 × 0 × 0.5) = 0

z = rcos∅ = 3(cos30°) = ( 3 × 0.866) = 2.6

∴ P.V of is OP₁ = [ 1.5i + 0j + 2.6k ]

P.V of P₂

x = rcos∅sin∅ = 3(cos45°) ( sin45°) = (3 × 0.7071 × 0.7071) = 1.5

y = rsin∅sin∅ = 3(sin45°) (sin45°) = (3 × 0.7071 × 0.7071) = 1.5

z = rcos∅ = 3(cos45°) = ( 3 × 0.7071) = 2.12

∴ P.V of is OP₂ = [ 1.5i + 1.5j + 2.12k ]

b)

Vector connecting P₁ to P₂ is given by

OP₂ - OP₁ = [ 1.5i + 1.5j + 2.12k ] - [ 1.5i + 0j + 2.6k ]

= [ 0i + 1.5j + 0.48k ]

c)

P₁P₂ → = [ 0i + 1.5j + 0.48k ] = [ 1.5j + 0.48k ]

so in a cylindrical coordinate, it should be

r = √(o² + 1.5²) = 1.5

∅ = tan⁻¹[y/π] = π/2

z = 0.48

cylindrical coordinates are (1.5, π/2, 0.48)

5 0

3 years ago