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elixir [45]
3 years ago
5

Help me with this question lol

Mathematics
1 answer:
romanna [79]3 years ago
5 0
Hope this helps.
Mark as brainliest

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A play was attended by 173 people. Adult tickets were $5 each and children paid $3 for each ticket the total revenue was $703. H
mr_godi [17]
The answer to this is 10545.
3 0
2 years ago
Read 2 more answers
1 2 3 4 5
mylen [45]

Answer:

B) 8x10^(-8)

Hope it helps

7 0
2 years ago
Plz help me with 13-16
svetlana [45]

Answer:

13) x = 10

14) x = 22

15) x = 12

16) x = -15

Step-by-step explanation:

13) 4x+5x=9x;90/9=10

14) 45-1=44;44/2=22

15) 37-1=36;36/3=12

16) -54+26=-28;x-28=3x+2;x=-15

6 0
2 years ago
2.- Supóngase que los diámetros de los tornillos fabricados por una compañía están distribuidos normalmente con una media de 0.2
Sergio [31]

Answer:

7.30167%

Step-by-step explanation:

Usando la fórmula de puntuación z

z = (x-μ) / σ, donde x es la puntuación bruta, μ es la media de la población y σ es la desviación estándar de la población

Para x <0.20 pulgadas

z = 0.20 - 0.25 / 0.02

z = -2.5

Valor de probabilidad de Z-Table:

P (x <0.20) = 0.0062097

Para x> 0.28 pulgadas

z = 0.28 - 0.20 / 0.02

z = 1.5

Valor de probabilidad de Z-Table:

P (x <0.28) = 0.93319

P (x> 0.28) = 1 - P (x <0.28) = 0.066807

La probabilidad de que se produzcan tornillos defectuosos cuando el tornillo se considera defectuoso si su diámetro es inferior a 0.20 pulgadas o superior a 0.28 pulgadas es

P (x <0.20) + P (x> 0.28)

= 0.0062097 + 0.066807

= 0.0730167

Conversión a porcentaje

= 0.0730167 × 100

= 7.30167%

El porcentaje de tornillos defectuosos producidos es

7.30167%

3 0
2 years ago
You are asked to do a study of shelters for abused and battered women to determine the necessary capacity in your city to provid
ohaa [14]

Answer:

Using the normal probability distribution, with a capacity of 350, it is enough for all abused on 90.82% of nights.

274 shelters will be needed.

Step-by-step explanation:

Problems of normally distributed samples are 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:

\mu = 250, \sigma = 75

If the city’s shelters have a capacity of 350, will that be enough places for abused women on 95% of all nights?

What is the percentile of 350?

This is the pvalue of Z when X = 350.

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

Z = \frac{250 - 150}{75}

Z = 1.33

Z = 1.33 has a pvalue of 0.9082.

Using the normal probability distribution, with a capacity of 350, it is enough for all abused on 90.82% of nights.

If not, what number of shelter openings will be needed?

The 95th percentile, which is X when Z = 1.645. So

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

1.645 = \frac{X - 150}{75}

X - 150 = 1.645*75

X = 274

274 shelters will be needed.

5 0
2 years ago
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