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2 years ago

Which number, when substituted for x, makes the equation 4x+8=164x+8=16 true?

Mathematics

2 answers:

ra1l [238]2 years ago

7 0

X = 2

4x + 8 = 16
-8 -8
----------------
4x= 8
-- --
4 4
------------
x=2

Guest

1 year ago

thanks dude

labwork [276]2 years ago

3 0

Answer:

$x=2$

Step-by-step explanation:

We have been given an equation $4x+8=16$ . we are asked to find the value of x that makes our given equation true.

Let us solve our given equation for x using opposite operations.

Subtract 8 from both sides:

$4x+8-8=16-8$

$4x=8$

Divide both sides by 4:

$\frac{4x}{4}=\frac{8}{4}$

$x=2$

Therefore, $x=2$ is the solution for our given equation.

Let us verify our answer:

$4x+8=16$

$4(2)+8=16$

$8+8=16$

$16=16$

Both sides are equal. Hence proven $x=2$ is solution for our given equation.

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Adrian Beltre hit 48 home runs during the 2004 Major League Baseball season, but only hit 19 home runs in the 2005 season. By wh

aliya0001 [1]

Answer:

60.42%

Step-by-step explanation:

Number of home runs hit by Adrian Beltre in 2004 = 48

Number of home runs hit by Adrian Beltre in 2005 = 19

To find:

Percentage decrease in home run production from 2004 to 2005.

Solution:

To find the percentage decrease, first of all we need to find the decrease in the number of home runs and then we will divide with the number of home runs in 2004 and then finally will multiply the result with 100 to get the percentage decrease.

Decrease in the number of home runs = Number of home runs in 2004 - Number of home runs in 2005 = 48 - 19 = 29

$\text{Percentage of decrease in Home runs} = \dfrac{\text{Decrease in home runs}}{\text{Number of home runs in 2004}}\times 100\\\Rightarrow \text{Percentage of decrease in Home runs} = \dfrac{29}{48}\times 100 \approx \bold{60.42\%}$

Therefore, by 60.42% Beltre's home run production has decreased from 2004 to 2005.

4 0

2 years ago

Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large un

Nataly_w [17]

Answer: 0.1289

Step-by-step explanation:

Given : The proportion of all students at a large university are absent on Mondays. : $p=0.15$

Sample size : $n=12$

Mean : $\mu=np=12\times0.15=1.8$

Standard deviation = $\sigma=\sqrt{np(1-p)}$

$\Rightarrow\ \sigma=\sqrt{12(0.15)(1-0.15)}=1.23693168769\approx1.2369$

Let x be a binomial variable.

Using the standard normal distribution table ,

$P(x=4)=P(x\leq4)-P(x\leq3)$ (1)

Z score fro normal distribution:-

$z=\dfrac{x-\mu}{\sigma}$

For x=4

$z=\dfrac{4-1.8}{1.2369}\approx1.78$

For x=3

$z=\dfrac{3-1.8}{1.2369}\approx0.97$

Then , from (1)

$P(x=4)=P(z\leq1.78)-P(z\leq0.97)\\\\=0.962462-0.8339768\approx0.1289$

Hence, the probability that four students are absent = $0.1289$

3 0

2 years ago

Answer plz. Zzzzxxzzz

Bezzdna [24]

Answer:

m= -12

Step-by-step explanation:

-15m-30=-12m+6

-3m=36

m=-12

7 0

1 year ago

Read 2 more answers

Write an expression for the calculation double 2 and then add5

Lyrx [107]

So,

"double 2"
2(2)

"add 5"
+ 5

Therefore, the whole expression would be:
2(2) + 5

If you wanted to evaluate it, it would come out like this.
4 + 5
9

5 0

2 years ago

One of the tallest buildings in a country is topped by a high antenna. The angle of elevation from the position of a surveyor on

irina1246 [14]

Answer:

a. distance of the surveyor to the base of the building = 2051.90 ft

b. height of the building = 1384 ft

c. Angle of elevation from the surveyor to the top of the antenna = 38.31°

d. Height of antenna = 237.08 ft

Step-by-step explanation:

The picture above is a illustration of the described event.

a = the height of the flag

b = the height of the building

c = distance of the surveyor from the base of the building

the angle of elevation from the position of the surveyor on the ground to the top of the building = 34°

distance from her position to the top of the building = 2475 ft

distance from her position to the top of the flag = 2615 ft

(a) How far away from the base of the building is the surveyor located?

using the SOHCAHTOA principle

cos 34° = c/2475

c = 0.8290375726 × 2475

c = 2051.8679921

c = 2051.90 ft

(b) How tall is the building

The height of the building = b

sin 34° = opposite /hypotenuse

0.5591929035 = b/2475

b = 0.5591929035 × 2475

b = 1384.0024361

b = 1384.00 ft

(c) What is the angle of elevation from the surveyor to the top of the antenna?

let the angle = ∅

cos ∅ = adjacent/hypotenuse

cos ∅ = 2051.90/2615

cos ∅ = 0.784665392

∅ = cos-1 0.784665392

∅ = 38.310258303

∅ = 38.31°

(d) How tall is the antenna?

height of the antenna = a

sin 38.31° = opposite/hypotenuse

sin 38.31° = (a + b)/2615

sin 38.31° × 2615 = (a + b)

(a + b) = 0.6199159917 × 2615

(a + b) = 1621.0803182

(a + b) = 1621. 08 ft

Height of antenna = 1621. 08 - 1384.00 = 237.08031822 ft

Height of antenna = 237.08 ft

8 0

2 years ago