Sheree Smith purchased a digital camera for $367.99. She received a manufacturer's $45 mail-in rebate and a

2. In the product of (4a3 + 5a2 – 11a) and (-15 + 3a – 7a2), the co-efficient of a4 is …………………… .

vekshin1

Answer:

- 23

Step-by-step explanation:

We could do this the long way and expand the whole product or note that the $a^{4}$ terms arise from the product of the a³ and a terms and the product of the a² and a² terms, that is

4a³ × 3a = 12 $a^{4}$

+ 5a² × - 7a² = - 35 $a^{4}$

Summing gives

12 $a^{4}$ - 35 $a^{4}$ = - 23 $a^{4}$

with coefficient - 23

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2 years ago

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A certain type of thread is manufactured with a mean tensile strength of 78.3 kilograms and a standard deviation of 5.6 kilogram

azamat

Answer:

(a) The variance decreases.

(b) The variance increases.

Step-by-step explanation:

According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the sample mean is given by,

$\mu_{\bar x}=\mu$

And the standard deviation of the sample mean is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}$

The standard deviation of sample mean is inversely proportional to the sample size, n.

So, if n increases then the standard deviation will decrease and vice-versa.

(a)

The sample size is increased from 64 to 196.

As mentioned above, if the sample size is increased then the standard deviation will decrease.

So, on increasing the value of n from 64 to 196, the standard deviation of the sample mean will decrease.

The standard deviation of the sample mean for n = 64 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{64}}=0.7$

The standard deviation of the sample mean for n = 196 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{196}}=0.4$

The standard deviation of the sample mean decreased from 0.7 to 0.4 when n is increased from 64 to 196.

Hence, the variance also decreases.

(b)

If the sample size is decreased then the standard deviation will increase.

So, on decreasing the value of n from 784 to 49, the standard deviation of the sample mean will increase.

The standard deviation of the sample mean for n = 784 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{784}}=0.2$

The standard deviation of the sample mean for n = 49 is:

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{5.6}{\sqrt{49}}=0.8$

The standard deviation of the sample mean increased from 0.2 to 0.8 when n is decreased from 784 to 49.

Hence, the variance also increases.

6 0

2 years ago

What is the product of (8.8 × 106)(5 × 102)?

Serhud [2]

(8.8 × 106)(5 × 102)

= (932.8)(510)

= 475,728

7 0

2 years ago

Emma earns $6 each time she mows the lawn and $8 per hour for babysitting. She is saving up to buy a new pair of jeans that cost

aleksandr82 [10.1K]

I would love to help but how can we help if we can not see the images ?

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2 years ago

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Starting from 300 feet away, a car drives toward you. It then passes by you at a speed of 48 feet per second. The distance dd (i

deff fn [24]

The answer to this question would be: 5 second

In this question, you are given the initial distance (300ft), car velocity(48ft/s) and the final distance(60ft). You are also given the equation to calculating the distance, so this question should be easier. The question is when the time where the distance is 60 ft. So the answer would be:

distance= 300-48t
60 = 300ft- 48t
48t= 300- 60=240
t= 5

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2 years ago

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