I need help with this

Find x <br><br> please help me out will mark you brainliest

Artemon [7]

Answer:

38+31+43+x=180

Step 1: Simplify both sides of the equation.

38+31+43+x=180

(x)+(38+31+43)=180(Combine Like Terms)

x+112=180

Step 2: Subtract 112 from both sides.

x+112−112=180−112

x=68

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

Use the interactive number line to find each difference to complete the table. A 4-column table with 4 rows. Column 1 is labeled

m_a_m_a [10]

Answer:

U=1

V=-5

W=-3

Step-by-step explanation:

3 0

3 years ago

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Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/m

borishaifa [10]

Answer:

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

Step-by-step explanation:

Air containing 0.04% carbon dioxide

V, volume of room is 6000 ft3.

Q, rate of air 2000 ft3/min,

initial concentration of 0.4% carbon dioxide,

determine the subsequent amount in the room at any time.

What is the concentration at 10 minutes?

firstly, we find the time taken for air to completely filled the room

Q = V/t

t = V/Q = 6000/2000 = 3min

so, its take 3mins for air to be completely filled in the room and for exhaust air to move out.

there is an initial concentration of 0.4% carbon dioxide, and the air pump in is 0.04%.

therefore,

3mins = 0.04% of CO2

3*60 =180sec = 0.04%

1sec = 0.04/180 = 0.00022%/sec

so at any time the concentration of CO2 is 0.4 + 0.00022 =0.40022%/sec

What is the concentration at 10 minute

the concentration at 10minutes = the concentration for 1minute because at every minutes, the concentration moves in is moves out. = concentration for 2000ft3.

for 0.04% = 6000ft3

? = 2000ft3

= 2000* 0.04)/6000 =0.0133%

the concentration at 10 minutes= 0.4+0.0133= 0.4133%

4 0

3 years ago

(t-distribution) A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. To m

statuscvo [17]

Answer:

There is an 38.21% probability that we find this lifespan for our sample average, or something even shorter.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

A manufacturing firm claims that the batteries used in laptop computers will last an average of 50 months. This means that $\mu = 50$ .

We found that the sample had a average lifespan of 47.3 months, and a standard deviation of s = 9 months. What is the probability that we find this lifespan for our sample average, or something even shorter?

We have to find the pvalue of Z when $X = 47.3$ .