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2 years ago

Zada needs to simplify the expression below. 72 – 4 × 12 ÷ 6 Which solution for Zada’s expression is correct?

Mathematics

1 answer:

BlackZzzverrR [31]2 years ago

8 0

Answer:

order of operations matter here so

72 - 4 × 12 ÷ 6

start with 4 × 12 (which equals 48)

72 - 48 ÷ 6

next is 48 ÷ 6. (which equals 8)

72 - 8 = 64

so g is correct

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9. A blueprint of a house plan uses a scale in

nata0808 [166]

Answer:

The length on the blueprint for an actual length of 65 feet is $16\frac{1}{4}\ inches$

Step-by-step explanation:

Given:

Scale of the blueprint is $\frac{1}{4}\ in=1\ foot$

Length of one side of the house is 65 feet.

Now, as per given data;

1 foot in actual = $\frac{1}{4}\ in$ on the blueprint.

Therefore, using unitary method, we can find find the length on the blueprint for an actual length of 65 feet by multiplying 65 and $\frac{1}{4}$ . Therefore,

Length on the blueprint for 65 feet is given as:

$=\frac{1}{4}\times 65\\\\=\frac{65}{4}\\\\=16\frac{1}{4}\ inches$

Therefore, the length on the blueprint for an actual length of 65 feet is $16\frac{1}{4}\ inches$

5 0

3 years ago

Find the following dates in the units given:

11Alexandr11 [23.1K]

Answer and step-by-step explanation:

a. 4 litres in 5 minutes (litres per minute)

4 litres per 5 minutes means 4/5 litres per 1 min = 0.8 litres per minutes

b. 12 litres in 45 minutes = 12/3 litres in 45/3 minutes = 4 litres per 15 minutes

60 minutes = 4*15 minutes

4 * 4 = 16 litres per hour

c. 800 kilowatt hours in 2 months

1 year = 12 months = 2*6 months

kilowatt per year = 800 * 6 = 4800 kilowatt per year

Hope this helped :3

3 0

2 years ago

It is known that the life of a particular auto transmission follows a normal distribution with mean 72,000 miles and standard de

scoray [572]

Answer:

a) $P(X$

$P(z$

b) $P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

$P(z>-0.583)=1-P(Z$

c) $P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

d) $z=1.28$

And if we solve for a we got

$a=72000 +1.28*12000=87360$

So the value of height that separates the bottom 90% of data from the top 10% is 87360.

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Part a

Let X the random variable that represent the life of a particular auto transmission of a population, and for this case we know the distribution for X is given by:

$X \sim N(72000,12000)$

Where $\mu=72000$ and $\sigma=12000$

We are interested on this probability

$P(X$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X$

And we can find this probability using excel or the normal standard table and we got:

$P(z$

Part b

$P(X>65000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>65000)=P(\frac{X-\mu}{\sigma}>\frac{65000-\mu}{\sigma})=P(Z>\frac{65000-72000}{12000})=P(z>-0.583)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>-0.583)=1-P(Z$

Part c

$P(X>100000)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X>100000)=P(\frac{X-\mu}{\sigma}>\frac{100000-\mu}{\sigma})=P(Z>\frac{100000-72000}{12000})=P(z>2.33)$

And we can find this probability using the complement rule and excel or the normal standard table and we got:

$P(z>2.33)=1-P(Z$

Sicne this probability just represent 1% of the data we can consider this value as unusual.

Part d

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X$