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3 years ago

6

in a box of thermometers 12% of them are broken and 66 are in good condition find a total number of the thermometers in the box.

1 answer:

Mama L [17]3 years ago

4 0

Answer:

75

Step-by-step explanation:

100% - 12% = 88% (% of good thermometers)

88% = 66

100% = 75 (total)

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A/20 + 14/15 = 9/15 <br><br> Can i get some help?

KiRa [710]

Method 1:

$\dfrac{a}{20}+\dfrac{14}{15}=\dfrac{9}{15}\ \ \ \ |multiply\ both\ sides\ by\ 60\\\\60\cdot\dfrac{a}{20}+60\cdot\dfrac{14}{15}=60\cdot\dfrac{9}{15}\\\\3a+4\cdot14=4\cdot9\\\\3a+56=36\ \ \ \ |subtract\ 56\ from\ both\ sides\\\\3a=-20\ \ \ \ |divide\ both\ sides\ by\ 3\\\\\boxed{a=-\frac{20}{3}\to a=-6\frac{2}{3}}$

Method 2.

$\dfrac{a}{20}+\dfrac{14}{15}=\dfrac{9}{15}\ \ \ \ |subtract\ \dfrac{14}{15}\ form\ both\ sides\\\\\dfrac{a}{20}=-\dfrac{5}{15}\\\\\dfrac{a}{20}=-\dfrac{1}{3}\ \ \ \ |multiply\ both\ sides\ by\ 20\\\\\boxed{a=-\dfrac{20}{3}\to a=-6\frac{2}{3}}$

3 0

3 years ago

The mean of a population is 74 and the standard deviation is 15. The shape of the population is unknown. Determine the probabili

Lena [83]

Answer:

a) 0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

b) 0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c) 0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

The mean of a population is 74 and the standard deviation is 15.

This means that $\mu = 74, \sigma = 15$

Question a:

Sample of 36 means that $n = 36, s = \frac{15}{\sqrt{36}} = 2.5$

This probability is 1 subtracted by the pvalue of Z when X = 78. So

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

By the Central Limit Theorem

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

$Z = \frac{78 - 74}{2.5}$

$Z = 1.6$

$Z = 1.6$ has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability of a random sample of size 36 yielding a sample mean of 78 or more.

Question b:

Sample of 150 means that $n = 150, s = \frac{15}{\sqrt{150}} = 1.2247$

This probability is the pvalue of Z when X = 77 subtracted by the pvalue of Z when X = 71. So

X = 77

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

$Z = \frac{77 - 74}{1.2274}$

$Z = 2.45$

$Z = 2.45$ has a pvalue of 0.9929

X = 71

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

$Z = \frac{71 - 74}{1.2274}$

$Z = -2.45$

$Z = -2.45$ has a pvalue of 0.0071

0.9929 - 0.0071 = 0.9858

0.9858 = 98.58% probability of a random sample of size 150 yielding a sample mean of between 71 and 77.

c. A random sample of size 219 yielding a sample mean of less than 74.2

Sample size of 219 means that $n = 219, s = \frac{15}{\sqrt{219}} = 1.0136$

This probability is the pvalue of Z when X = 74.2. So

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

$Z = \frac{74.2 - 74}{1.0136}$

$Z = 0.2$

$Z = 0.2$ has a pvalue of 0.5793

0.5793 = 57.93% probability of a random sample of size 219 yielding a sample mean of less than 74.2

5 0

3 years ago

Today, both the soccer team and the basket ball team had games (starting on Monday). The soccer team plays every 3 days and the

yarga [219]

I think eight because if you add those days together they will play again in eight days so that they play on the same day.

Sorry if I got it wrong I tried and that's what counts, right?

3 0

3 years ago

During 4 days, the price of the stock of PEV Corporation went up 1/4 of a point, down 1/3 of a point, down 3/4 of a point, and u

HACTEHA [7]

The price of the stock of PEV Corporation varies as shown. we have to determine the net charge. Net change is just how much it changed overall, so we will look for all the ups minus all the downs.

Here, "went up" goes with a + sign, and "went down" goes with a - sign.

So, Net charge = $\frac{1}{4} - \frac{1}{3}-\frac{3}{4} + \frac{7}{10}$

= $\frac{1}{4}-\frac{3}{4}- \frac{1}{3} + \frac{7}{10}$

= $-\frac{2}{4}- \frac{1}{3} + \frac{7}{10}$

= $-\frac{1}{2}- \frac{1}{3} + \frac{7}{10}$

LCM of 2,3 and 10 is 30

= $\frac{-15-10+21}{30}$

= $\frac{-4}{30}$

= $\frac{-2}{15}$

So, the net charge is $\frac{-2}{15}$

8 0

3 years ago

Find the least common multiple of 10,15,25 please no links or photo

sergey [27]

Answer:

150

Step-by-step explanation:

The LCM of 10,15,25 10 , 15 , 25 is 2⋅3⋅5⋅5=150 2 ⋅ 3 ⋅ 5 ⋅ 5 = 150 .

6 0

3 years ago