All categories

3 years ago

Solve 8(1/4×+3/4)-3=17

2 answers:

7 0

8(1/4x + 3/4) -3 =17

First distribute which results in (8/4x + 24/4)-3 =17.
Now just simply simplify it into (2x+6)-3 =17
Then you combine like term : 2x+3 =17

Subtract 3 on both sides : 2x=14

Finally divide 2 on both sides to isolate x: x=7

marta [7]3 years ago

4 0

Answer:

x=7

Step-by-step explanation:

8(1/4x+3/4)-3=17

Add 3 to each side

8(1/4x+3/4)-3+3=17+3

8(1/4x+3/4)=20

Distribute the 8

2x +6 =20

Subtract 6 from each side

2x+6-6 = 20-6

2x=14

Divide by 2

2x/2 =14/2

x=7

You might be interested in

This are 3 different questions please help me

Marysya12 [62]

1. diginity
2.require
3.tuition

5 0

2 years ago

A sample size 25 is picked up at random from a population which is normally

Margarita [4]

Answer:

a) P(X < 99) = 0.2033.

b) P(98 < X < 100) = 0.4525

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.

Mean of 100 and variance of 36.

This means that $\mu = 100, \sigma = \sqrt{36} = 6$

Sample of 25:

This means that $n = 25, s = \frac{6}{\sqrt{25}} = 1.2$

(a) P(X<99)

This is the pvalue of Z when X = 99. So

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

By the Central Limit Theorem

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

$Z = \frac{99 - 100}{1.2}$

$Z = -0.83$

$Z = -0.83$ has a pvalue of 0.2033. So

P(X < 99) = 0.2033.

b) P(98 < X < 100)

This is the pvalue of Z when X = 100 subtracted by the pvalue of Z when X = 98. So

X = 100

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

$Z = \frac{100 - 100}{1.2}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

X = 98

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

$Z = \frac{98 - 100}{1.2}$

$Z = -1.67$

$Z = -1.67$ has a pvalue of 0.0475

0.5 - 0.0475 = 0.4525

P(98 < X < 100) = 0.4525

6 0

2 years ago

Jace went shopping for a new phone. The listed price of the phone was $50, but the

sukhopar [10]

Answer:

1.50

Step-by-step explanation:

3 0

2 years ago

What is the following sum ?

Hitman42 [59]

Answer: 4th option

Explanation:

I added a picture so please check it

6 0

2 years ago

Read 2 more answers

Because of staffing decisions, managers of the Gibson-Marimont Hotel are interested in

Law Incorporation [45]

Answer:

(a) 900

(b) [567.35 , 1689.72]

(c) [23.82 , 41.11]

Step-by-step explanation:

We are given that a sample of 20 days of operation shows a sample mean of 290 rooms occupied per day and a sample standard deviation of 30 rooms i.e.;

Sample mean, $xbar$ = 290 Sample standard deviation, s = 30 and Sample size, n = 20

(a) Point estimate of the population variance is equal to sample variance, which is the square of Sample standard deviation ;

$\sigma^{2}$ = $s^{2}$ = $30^{2}$

$\sigma^{2}$ = 900

(b) 90% confidence interval estimate of the population variance is given by the pivotal quantity of $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

P(10.12 < $\chi^{2}__1_9$ < 30.14) = 0.90 {At 10% significance level chi square has critical

values of 10.12 and 30.14 at 19 degree of freedom}

P(10.12 < $\frac{(n-1)s^{2} }{\sigma^{2} }$ < 30.14) = 0.90

P( $\frac{10.12}{(n-1)s^{2} }$ < $\frac{1 }{\sigma^{2} }$ < $\frac{30.14}{(n-1)s^{2} }$ ) = 0.90

P( $\frac{(n-1)s^{2} }{30.14}$ < $\sigma^{2}$ < $\frac{(n-1)s^{2} }{10.12}$ ) = 0.90

90% confidence interval for $\sigma^{2}$ = $[\frac{19s^{2} }{30.14} , \frac{19s^{2} }{10.12}]$

= $[\frac{19*900 }{30.14} , \frac{19*900 }{10.12}]$

= [567.35 , 1689.72]

Therefore, 90% confidence interval estimate of the population variance is [567.35 , 1689.72] .

P( $\sqrt{\frac{(n-1)s^{2} }{30.14}}$ < $\sigma$ < $\sqrt{\frac{(n-1)s^{2} }{10.12}}$ ) = 0.90

90% confidence interval for $\sigma$ = $[\sqrt{\frac{19s^{2} }{30.14}} , \sqrt{\frac{19s^{2} }{10.12}} ]$

= [23.82 , 41.11]

Therefore, 90% confidence interval estimate of the population standard deviation is [23.82 , 41.11] .

7 0

3 years ago