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3 years ago

Two times a number added to -4, subtracted from triple the sum of eight times the number and five

Mathematics

2 answers:

mr Goodwill [35]3 years ago

7 0

Answer:

2x+(-4)-8x(3)-5

lapo4ka [179]3 years ago

5 0

Answer:

-22a-9

Step-by-step explanation:

2a+(-4)-3(8a+5)

2a+4-24a-15

2a-24a+4-15

-22a-9

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Tanya took a cab from her home to the airport. Her total fare including a tip of $5.00, was $28.50. EZ Cab Company has a pick-up

ValentinkaMS [17]

16.2 miles.

Since she paid a $5 tip, the cost for her fare was $23.50. Let x be the number of 0.2-mile increments in the trip:
23.50 = 2.50 + 1 + 0.25(x-1)

This equation comes from the 2.50 home pick-up fee; $1 for the first 0.2-mile increment; and 0.25 for each other 0.2-mile increment.

Using the distributive property, we have:
23.50 = 2.50 + 1 + 0.25*x - 0.25*1
23.50 = 2.50 + 1 + 0.25x - 0.25

Combining like terms,
23.50 = 0.25x + 2.75

Subtracting 2.75 from both sides, we have:
23.50 - 2.75 = 0.25x + 2.75 - 2.75
20.25 = 0.25x

Dividing both sides by 0.25, we have:
20.25/0.25 = 0.25x/0.25
81 = x

This means there are 81 0.2-mile increments in her trip. 81(0.2) = 16.2 miles

6 0

4 years ago

WHO IS REALLY GOOD AT MATH AND CAN DO THIS FOR ME ILL MARK BRAINLIEST

Rashid [163]

Answer:

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

PLEASE HELP I WILL GIVE BRAINLEST

Ira Lisetskai [31]

The theoretical probability of getting heads or tails is 1/2 the time.

1/2 of 150 = 75

So theoretically you would get 75 heads and 75 tails.

The experiment got 84 heads, which is greater than 75, so it would be higher than the theoretical probability.

The answer is : It is 6% higher than the theoretical probability.

4 0

3 years ago

An industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specifica

Natalija [7]

Answer:

(a) Probability that a ball bearing is between the target and the actual mean is 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is 0.226.

(c) Probability that a ball bearing is above the upper specification limit is 0.0401.

(d) Probability that a ball bearing is below the lower specification limit is 0.0006.

Step-by-step explanation:

We are given that an industrial sewing machine uses ball bearings that are targeted to have a diameter of 0.75 inch. The lower and upper specification limits under which the ball bearings can operate are 0.74 inch and 0.76 inch, respectively.

Past experience has indicated that the actual diameter of the ball bearings is approximately normally distributed, with a mean of 0.753 inch and a standard deviation of 0.004 inch.

Let X = diameter of the ball bearings

SO, X ~ Normal( $\mu=0.753,\sigma^{2} =0.004^{2}$ )

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

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

where, $\mu$ = population mean = 0.753 inch

$\sigma$ = standard deviation = 0.004 inch

(a) Probability that a ball bearing is between the target and the actual mean is given by = P(0.75 < X < 0.753) = P(X < 0.753 inch) - P(X $\leq$ 0.75 inch)

P(X < 0.753) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.753-0.753}{0.004} } }$ ) = P(Z < 0) = 0.50

P(X $\leq$ 0.75) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.75-0.753}{0.004} } }$ ) = P(Z $\leq$ -0.75) = 1 - P(Z < 0.75)

= 1 - 0.7734 = 0.2266

The above probability is calculated by looking at the value of x = 0 and x = 0.75 in the z table which has an area of 0.50 and 0.7734 respectively.

Therefore, P(0.75 inch < X < 0.753 inch) = 0.50 - 0.2266 = 0.2734.

(b) Probability that a ball bearing is between the lower specification limit and the target is given by = P(0.74 < X < 0.75) = P(X < 0.75 inch) - P(X $\leq$ 0.74 inch)

P(X < 0.75) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.75-0.753}{0.004} } }$ ) = P(Z < -0.75) = 1 - P(Z $\leq$ 0.75)

= 1 - 0.7734 = 0.2266

P(X $\leq$ 0.74) = P( $\frac{X-\mu}{\sigma} } }$ $\leq$ $\frac{0.74-0.753}{0.004} } }$ ) = P(Z $\leq$ -3.25) = 1 - P(Z < 3.25)

= 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 0.75 and x = 3.25 in the z table which has an area of 0.7734 and 0.9994 respectively.

Therefore, P(0.74 inch < X < 0.75 inch) = 0.2266 - 0.0006 = 0.226.

(c) Probability that a ball bearing is above the upper specification limit is given by = P(X > 0.76 inch)

P(X > 0.76) = P( $\frac{X-\mu}{\sigma} } }$ > $\frac{0.76-0.753}{0.004} } }$ ) = P(Z > -1.75) = 1 - P(Z $\leq$ 1.75)

= 1 - 0.95994 = 0.0401

The above probability is calculated by looking at the value of x = 1.75 in the z table which has an area of 0.95994.

(d) Probability that a ball bearing is below the lower specification limit is given by = P(X < 0.74 inch)

P(X < 0.74) = P( $\frac{X-\mu}{\sigma} } }$ < $\frac{0.74-0.753}{0.004} } }$ ) = P(Z < -3.25) = 1 - P(Z $\leq$ 3.25)

= 1 - 0.9994 = 0.0006

The above probability is calculated by looking at the value of x = 3.25 in the z table which has an area of 0.9994.

8 0

4 years ago

Becky purchased a home entertainment center set for $2254

vlabodo [156]

Wheres the answer choices

5 0

3 years ago

Read 2 more answers