Solve log2 x = log5 3+1 by graphing. round to the nearest tenth

Lashonda sells beaded necklaces. Each large necklace sells for $6.60 and each small necklace sells for $4.90 . How much will she

xxMikexx [17]

46.2 dollars for the large necklaces and 14 dollars and seventy cents for the small necklaces therefore she will earn $60.9

5 0

3 years ago

Calculate the average speed of a train that

lorasvet [3.4K]

Answer:

4

Step-by-step explanation:

4 0

3 years ago

Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%. Find

Papessa [141]

Answer:

1) $\text{P(at least one boy and one girl)}=\frac{3}{4}$

2) $\text{P(at least one boy and one girl)}=\frac{3}{8}$

3) $\text{P(at least two girls)}=\frac{1}{2}$

Step-by-step explanation:

Given : Mr. and Mrs. Romero are expecting triplets. Suppose the chance of each child being a boy is 50% and of being a girl is 50%.

To Find : The probability of each event.

1) P(at least one boy and one girl)

2) P(two boys and one girl)

3) P(at least two girls)

Solution :

Let's represent a boy with B and a girl with G

Mr. and Mrs. Romero are expecting triplets.

The possibility of having triplet is

BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

Total outcome = 8

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

1) P(at least one boy and one girl)

Favorable outcome = BBG, BGB, BGG, GBB, GBG, GGB=6

$\text{P(at least one boy and one girl)}=\frac{6}{8}$

$\text{P(at least one boy and one girl)}=\frac{3}{4}$

2) P(at least one boy and one girl)

Favorable outcome = BBG, BGB, GBB=3

$\text{P(at least one boy and one girl)}=\frac{3}{8}$

3) P(at least two girls)

Favorable outcome = BGG, GBG, GGB, GGG=4

$\text{P(at least two girls)}=\frac{4}{8}$

$\text{P(at least two girls)}=\frac{1}{2}$

4 0

3 years ago

Read 2 more answers

X – 6 = 14 – 8 12 0 20 –12

sineoko [7]

Hi there!

Answer:

A. 12. x=12 and 12=x

Step-by-step explanation:

addition property of equality is adding the same number both sides of an equation does not change the equation.

subtraction property of equality is subtraction the same number from both sides of an equation does not change the equation.

multiplication property of equality is multiplying both sides of an equation by the same number does not change the equation.

division property of equality is dividing both sides of an equation by the same number non-zero number does not change the equation.

x-6=14-8

subtract by 6 both sides of an equation.

x-6=6

add by 6 both sides of an equation.

x-6+6=6+6

simplify.

6+6=12

x=12

12-6=6

Hope this helps!

-Charlie

Have a great day!

7 0

3 years ago

what is the surface area of a conical grain storage tank that has a height of 24 meters and a diameter of 16 meters? round the a

Sauron [17]

The basis of this cone is a circle with a diameter d = 16m. This information is sufficient to calculate
1. The surface of the wheel . We use the formula

 $A = \pi r^{2}$

2. the radius of the wheel needed to the formula for calculating the area of a circle and the cone

d = 16m
1/2d = r = 1/2*16m = 8m
..................................................................................................
Area base
 $P = \pi r^{2} = \pi (8m)^{2} = 64 \pi m^{2}$
..................................................................................................
The surface area

$A = \pi rl$
l - creates a cone = ?

Height , forming a cone and the radius of a circle formed triangle . Two known sides is possible to calculate the third side ( the cone ) of the Pythagorean Theorem

a = r = 8cm
b = h = 24m
c = l = ?

a² + b² = c²
r² + h² = l²
l = √ (r² + h²) = √ [ ( 8m)² + ( 24m)² ] = √ (64m² + 576m²) = √ (640 m²)= √ (8*8*10m²) = 8√10m

The lateral surface of the cone
A = πrl = π*8m *8√10m = 64√10πm²

The total area of the cone = The base area of the side surface of +

A = 64√10πm² + 64πm² = 64πm² *(√10 + 1)

3 0

3 years ago