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2 years ago

Evaluate the numerical expression 2+65 + 7*3=

Mathematics

2 answers:

riadik2000 [5.3K]2 years ago

7 0

2 + 65 + 7 × 3
= 2 + 65 + 21
= 67 + 21
= 88

answer is 88

Leya [2.2K]2 years ago

5 0

This is a very simple equation however, you need to know the order of operations. First do 7x3 which is 21 then add it to 65+2 to get....::67+21=88 YAYAYAYAY

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The cost of a company picnic varies directly as the number of employees attending the picnic. If a company picnic costs $487.50

Drupady [299]

The company picnic cost $ 1300 for 80 employees

Solution:

Given that cost of a company picnic varies directly as the number of employees attending the picnic

Let "c" be the company picnic cost

Let "n" be the number of employees attending the picnic

Therefore,

$cost \propto \text{ number of employees attending picnic }$

$c \propto n\\\\c = kn$

Where "k" is the constant of proportionality

c = kn ---------- eqn 1

Given that company picnic costs $487.50 for 30 employees

Therefore substitute c = 487.50 and n = 30

$487.50 = k(30)\\\\k = \frac{487.50}{30} = 16.25$

How much does a company picnic cost for 80 employees?

Substitute n = 80 and k = 16.25 in eqn 1

$c = 16.25(80) = 1300$

Thus $ 1300 is the cost for 80 employees

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3 years ago

A pharmaceutical company receives large shipments of aspirin tablets. The acceptance sampling plan is to randomly select and tes

DerKrebs [107]

Answer:

a) The probability that this whole shipment will be accepted is 30%.

b) Many of the shipments with this rate of defective aspirin tablets will be rejected.

Step-by-step explanation:

We have a shipment of 3000 aspirin tablets, with a 5% rate of defects.

We select a sample of size 48 and test for defectives.

If more than one aspirin is defective, the batch is rejected.

The amount of defective aspirin tablets X can be modeled as a binomial distribution random variable, with p=0.55 and n=48

We have to calculate the probabilities that X is equal or less than 1: P(X≤1).

$P(X\leq1)=P(X=0)+P(X=1)\\\\\\P(0)=\binom{48}{0}(0.05)^0(0.95)^{48}=1*1*0.0853=0.0853\\\\\\P(1)=\binom{48}{1}(0.05)^1(0.95)^{47}=48*0.05*0.0897=0.2154\\\\\\P(X\eq1)=0.0853+0.2154=0.3007$

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3 years ago

This is due tm please help

Mumz [18]

Answer:

25. No slope

Step-by-step explanation:

25. Y2 - Y1 / X2 - X1

= -8 - (-3) / 9 - 9

= -5 / 0

= no slope; straight line

Use this formula (Y2 - Y1 / X2 - X1) for each ques.

8 0

3 years ago

Suppose a geyser has a mean time between irruption’s of 75 minutes. If the interval of time between the eruption is normally dis

lesya [120]

Answer:

(a) The probability that a randomly selected Time interval between irruption is longer than 84 minutes is 0.3264.

(b) The probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is 0.0526.

(c) The probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is 0.0222.

(d) The probability decreases because the variability in the sample mean decreases as we increase the sample size

(e) The population mean may be larger than 75 minutes between irruption.

Step-by-step explanation:

We are given that a geyser has a mean time between irruption of 75 minutes. Also, the interval of time between the eruption is normally distributed with a standard deviation of 20 minutes.

(a) Let X = the interval of time between the eruption

So, X ~ Normal( $\mu=75, \sigma^{2} =20$ )

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

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

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

Now, the probability that a randomly selected Time interval between irruption is longer than 84 minutes is given by = P(X > 84 min)

P(X > 84 min) = P( $\frac{X-\mu}{\sigma}$ > $\frac{84-75}{20}$ ) = P(Z > 0.45) = 1 - P(Z $\leq$ 0.45)

= 1 - 0.6736 = 0.3264

The above probability is calculated by looking at the value of x = 0.45 in the z table which has an area of 0.6736.

(b) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

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

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 13

Now, the probability that a random sample of 13 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{13} } }$ ) = P(Z > 1.62) = 1 - P(Z $\leq$ 1.62)

= 1 - 0.9474 = 0.0526

The above probability is calculated by looking at the value of x = 1.62 in the z table which has an area of 0.9474.

(c) Let $\bar X$ = sample time intervals between the eruption

The z-score probability distribution for the sample mean is given by;

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

where, $\mu$ = population mean time between irruption = 75 minutes

$\sigma$ = standard deviation = 20 minutes

n = sample of time intervals = 20

Now, the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is given by = P( $\bar X$ > 84 min)

P( $\bar X$ > 84 min) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{84-75}{\frac{20}{\sqrt{20} } }$ ) = P(Z > 2.01) = 1 - P(Z $\leq$ 2.01)

= 1 - 0.9778 = 0.0222

The above probability is calculated by looking at the value of x = 2.01 in the z table which has an area of 0.9778.

(d) When increasing the sample size, the probability decreases because the variability in the sample mean decreases as we increase the sample size which we can clearly see in part (b) and (c) of the question.

(e) Since it is clear that the probability that a random sample of 20 time intervals between irruption has a mean longer than 84 minutes is very slow(less than 5%0 which means that this is an unusual event. So, we can conclude that the population mean may be larger than 75 minutes between irruption.

8 0

3 years ago

Traveling at 65 mph how many feet can you travel in 22 minutes?

baherus [9]

If you are travelling at 65 mph, you have to think of it as if you are moving at a constant speed. By dividing 65 by 60, you are calculating how many miles you are moving each minute. In doing so, you can calculate that you are moving at 1.083333 miles each minute. That's approximately 5720 feet per minute. (This is calculated by multiplying 1.083333 by 5280 (the number of feet in a mile)). 5720 x 22 minutes, you will have traveled 125,840 feet.

Hope this helps!

6 0

3 years ago

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