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3 years ago

Solve for x: 9(x+1) = 25+X

Mathematics

2 answers:

Helga [31]3 years ago

6 0

Answer:

$x = 2$

Step-by-step explanation:

Step 1: Distribute

$9(x + 1) = 25 + x$

$(9 * x) + (9 * 1) = 25 + x$

$9x + 9 = 25 + x$

Step 2: Subtract x from both sides

$9x + 9 - x = 25 + x - x$

$8x + 9 = 25$

Step 3: Subtract 9 from both sides

$8x + 9 - 9 = 25 - 9$

$8x = 16$

Step 4: Divide both sides by 8

$8x / 8 = 16 / 8$

$x = 2$

Answer: $x = 2$

Inessa [10]3 years ago

5 0

Answer:

$x=2$

Step-by-step explanation:

$9(x+1) = 25+x$

Distribute the 9:

$9x+9=25+x$

Subtract x on both sides:

$8x+9=25$

Subtract 9 on both sides:

$8x=16$

Divide both sides by 8:

$x=2$

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Step-by-step explanation:

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Using the addition or subtraction formulas for sine or cosine.... sinasinb=(cos(a-b)-cos(a+b))/2

raketka [301]

Answer:

The answer w'll be obtained using formulas

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cos(a-b) = cosacosb + sinasinb

Step-by-step explanation:

Using the trigonometric formula of addition and subtraction of cosine

cos(a+b) = cosacosb - sinasinb

cos(a-b) = cosacosb + sinasinb

w'll get the desired answer.

To be solve

L.H.S = R.H.S

sinasinb = (cos(a-b)-cos(a+b)/2

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sinasinb = (cos(a-b) - (cosacosb -sinasinb))/2

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3 years ago

Automated manufacturing operations are quite precise but still vary, often with distribution that are close to Normal. The width

konstantin123 [22]

Answer:

The proportion of slots which meet these specifications is 0.16634 or 16.63%.

Step-by-step explanation:

We are given that the width in inches of slots cut by a milling machine follows approximately the N(0.72,0.0012) distribution.

Also, the specifications allow slot widths between 0.71975 and 0.72025.

Let X = width in inches of slots cut by a milling machine

The z-score probability distribution for normal distribution is given by;

Z = $\frac{ X-\mu}{{\sigma}} }} }$ ~ N(0,1)

where, $\mu$ = population mean width = 0.72

$\sigma$ = standard deviation = 0.0012

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the specifications allow slot widths between 0.71975 and 0.72025 is given by = P(0.71975 < X < 0.72025)

P(0.71975 < X < 0.72025) = P(X < 0.72025) - P(X $\leq$ 0.71975)

P(X < 0.72025) = P( $\frac{ X-\mu}{{\sigma}} }} }$ < $\frac{ 0.72025-0.72}{{0.0012}} }} }$ ) = P(Z < 0.21) = 0.58317

P(X $\leq$ 0.71975) = P( $\frac{ X-\mu}{{\sigma}} }} }$ $\leq$ $\frac{ 0.71975-0.72}{{0.0012}} }} }$ ) = P(Z $\leq$ -0.21) = 1 - P(Z < 0.21)

= 1 - 0.58317 = 0.41683

So, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.21 in the z table which has an area of 0.58317.

Therefore,  P(0.71975 < X < 0.72025) = 0.58317 - 0.41683 = 0.16634

Hence, the proportion of slots who meet these specifications is 16.63%.

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3 years ago