- 4(5x - 1) – 8x+11 + -17-3(2x + 13) + 24 simplify

-5x– 8< -53 and -63 < -5x – 8

baherus [9]

Answer:

x>9 and x<11

Step-by-step explanation:

7 0

2 years ago

The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2617 and standard deviation

Eduardwww [97]

Answer:

a) N(2617, 586)

b) 0.3613 = 36.13% probability that the customer consumes less than 2409 calories.

c) 0.4013 = 40.13% of the customers consume over 2764 calories

d) 3981 calories.

Step-by-step explanation:

Problems of normally distributed samples can be 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 question, we have that:

$\mu = 2617, \sigma = 586$

a. What is the distribution of X?

Here we first place the mean, then the standard deviation.

N(2617, 586)

b. Find the probability that the customer consumes less than 2409 calories.

This is the pvalue of Z when X = 2409. So

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

$Z = \frac{2409 - 2617}{586}$

$Z = -0.355$

$Z = -0.355$ has a pvalue of 0.3613

0.3613 = 36.13% probability that the customer consumes less than 2409 calories.

c. What proportion of the customers consume over 2764 calories?

This is 1 subtracted by the pvalue of Z when X = 2764. So

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

$Z = \frac{2764 - 2617}{586}$

$Z = 0.25$

$Z = 0.25$ has a pvalue of 0.5987

1 - 0.5987 = 0.4013

0.4013 = 40.13% of the customers consume over 2764 calories

d. The Piggy award will given out to the 1% of customers who consume the most calories. What is the fewest number of calories a person must consume to receive the Piggy award?

Top 1%, so the 100-1 = 99th percentile.

The 99th percentile is the value of X when Z has a pvalue of 0.99. So it is X when Z = 2.327. So

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

$2.327 = \frac{X - 2617}{586}$

$X - 2617 = 2.327*586$

$X = 3980.6$

Rounding to the nearest calorie, 3981 calories.

7 0

2 years ago

Hi hi um help. i'm too tired to think.

leva [86]

Answer:

I think 66?

I hope it is if it’s not I’m sorry

Step-by-step explanation:

5 0

2 years ago

Read 2 more answers

Cable Strength: A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the ca

KatRina [158]

Answer:

95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].

Step-by-step explanation:

We are given that the engineers take a random sample of 45 cables and apply weights to each of them until they break. The mean breaking weight for the 45 cables is 768.2 lb. The standard deviation of the breaking weight for the sample is 15.1 lb.

Since, in the question it is not specified that how much confidence interval has be constructed; so we assume to be constructing of 95% confidence interval.

Firstly, the Pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean breaking weight = 768.2 lb

s = sample standard deviation = 15.1 lb

n = sample of cables = 45

$\mu$ = population mean breaking strength

Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.02 < $t_4_4$ < 2.02) = 0.95 {As the critical value of t at 44 degree

of freedom are -2.02 & 2.02 with P = 2.5%}

P(-2.02 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.02) = 0.95

P( $-2.02 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $2.02 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.02 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.02 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.02 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.02 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $768.2-2.02 \times {\frac{15.1}{\sqrt{45} } }$ , $768.2+2.02 \times {\frac{15.1}{\sqrt{45} } }$ ]

= [763.65 lb , 772.75 lb]

Therefore, 95% confidence interval for the mean breaking strength of the new steel cable is [763.65 lb , 772.75 lb].

3 0

3 years ago

Bonjour quelqu'un peut m'aider? Transformer les sommes ou les differences suivantes en un produit : A= 100* n - 2* n ; B= 14*30

BartSMP [9]

Answer:

J’espère que cela vous aidera;

A= 98n; B = 630; C = 22t; D = 510;E = 25m ; F = 335

Step-by-step explanation:

3 0

2 years ago

-4(5x - 1) – 8x+11 + -17-3(2x + 13) + 24simplify​

-4(5x - 1) – 8x+11 + -17-3(2x + 13) + 24simplify