Please help me with this surds question.

Five bags of chips cost $7.

Dvinal [7]

Answer:

1. $1.40 2. $8.40

Step-by-step explanation:

1. 7 divided by 5 is 1.4

2. 1.40 plus $7 equals $8.40

6 0

2 years ago

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What is the best way to Solve: 25 – 5x = 80

Art [367]

25 - 5x = 80
+ 5x + 5x

25 = 5x + 80
- 80 -80

-55 = 5x

-55/5 = 11

x = 11

Hope I helped!

Let me know if you need anything else!

~ Zoe

5 0

3 years ago

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Someone help me pls show work

padilas [110]

Answer:

*insert answer*

Step-by-step explanation:

I dont know it but hey

:D

8 0

2 years ago

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Big chickens: The weights of broilers (commercially raised chickens) are approximately normally distributed with mean 1387 grams

Nataliya [291]

Answer:

a) 0.2318

b) 0.2609

c) No it is not unusual for a broiler to weigh more than 1610 grams

Step-by-step explanation:

We solve using z score formula

z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean 1387 grams and standard deviation 192 grams. Use the TI-84 Plus calculator to answer the following.

(a) What proportion of broilers weigh between 1150 and 1308 grams?

For 1150 grams

z = 1150 - 1387/192

= -1.23438

Probability value from Z-Table:

P(x = 1150) = 0.10853

For 1308 grams

z = 1308 - 1387/192

= -0.41146

Probability value from Z-Table:

P(x = 1308) = 0.34037

Proportion of broilers weigh between 1150 and 1308 grams is:

P(x = 1308) - P(x = 1150)

0.34037 - 0.10853

= 0.23184

≈ 0.2318

(b) What is the probability that a randomly selected broiler weighs more than 1510 grams?

1510 - 1387/192

= 0.64063

Probabilty value from Z-Table:

P(x<1510) = 0.73912

P(x>1510) = 1 - P(x<1510) = 0.26088

≈ 0.2609

(c) Is it unusual for a broiler to weigh more than 1610 grams?

1610- 1387/192

= 1.16146

Probability value from Z-Table:

P(x<1610) = 0.87727

P(x>1610) = 1 - P(x<1610) = 0.12273

≈ 0.1227

No it is not unusual for a broiler to weigh more than 1610 grams

8 0

3 years ago

A toy manufacturer wants to know how many new toys children buy each year. A sample of 305 children was taken to study their pur

Harrizon [31]

Answer:

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.8}{2} = 0.1$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of $1 - 0.1 = 0.9$ , so Z = 1.28.

Now, find the margin of error M as such

$M = z\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.28\frac{1.5}{\sqrt{305}} = 0.1$

The lower end of the interval is the sample mean subtracted by M. So it is 7.6 - 0.1 = 7.5

The upper end of the interval is the sample mean added to M. So it is 7.6 + 0.1 = 7.7

The 80% confidence interval for the mean number of toys purchased each year is between 7.5 and 7.7 toys.

8 0

3 years ago