All categories

3 years ago

Make n the subject of the formula: M=3n

Mathematics

1 answer:

natta225 [31]3 years ago

4 0

Divide each side by 3. ----- n=M/3 .

You might be interested in

15% of R560 - 15% of R500 is:

Dahasolnce [82]

15% of R560 - 15% of R500

=> 0.15 × 560 – 0.15 × 500

=> 0.15 ( 560–500)

=> 0.15 × 60

=> Rs. 9

HOPE IT HELPS

PLEASE MARK ME BRAINLIEST ☺️

8 0

2 years ago

A car travels 17 miles on one gallon of gas

zmey [24]

When x =170, y =10, when x = 187, y = 1, when x = 204, y = 12, when x = 221, y = 13, when x = 238, y = 14, when x = 255, y = 15, when x = 272, y = 16, when x = 289, y = 17, when x = 306, y = 18, when x = 323, y = 19, when x = 340, y = 20,

Step-by-step explanation:

Step 1; Assume y is the gallons in the tank while x is the range of the car. We need to determine the equation relating the y to the x values in the given coordinates. x is the dependent value whereas y is the independent value i.e. the value of x depends on the value of y.

Step 2; As y increases, x also increases. For every extra gallon in the tank, the car can go another 17 miles. So with y as the amount of gas in the tank in gallons and x as the range of vehicle, we get;

x = 17y, this represents the relation between x and y.

To find the gallons in the tank, y = $\frac{x}{17}$ ,

For x = 170, y = $\frac{170}{17}$ = 10,

Similarly equating the values we get that when x = 187, y = 11,

when x = 204, y = 12,

when x = 221, y = 13,

when x = 238, y = 14,

when x = 255, y = 15,

when x = 272, y = 16,

when x = 289, y = 17,

when x = 306, y = 18,

when x = 323, y = 19,

when x = 340, y = 20,

8 0

2 years ago

The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in

Stolb23 [73]

Answer:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

Step-by-step explanation:

Assuming this complete question: The Wisconsin Dairy Association is interested in estimating the mean weekly consumption of milk for adults over the age of 18 in that state. To do this, they have selected a random sample of 300 people from the designated population. The following results were recorded: xbar=34.5 ounces, s=7.9 ounces Given this information, if the leaders wish to estimate the mean milk consumption with 90 percent confidence, what is the approximate margin of error in the estimate?

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

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".

$\bar X=34.5$ represent the sample mean

$\mu$ population mean (variable of interest)

s=7.9 represent the sample standard deviation

n=300 represent the sample size

Solution to the problem

The confidence interval for the mean is given by the following formula:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=300-1=299$

Since the Confidence is 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,199)".And we see that $t_{\alpha/2}=1.653$

And the margin of error is given by:

$ME = 1.653 *\frac{7.9}{\sqrt{300}}= \pm 0.75$

±0.75 ounce

4 0

3 years ago

Read 2 more answers

The U.S. government has devoted considerable funding to missile defense research over the past 20 years. The latest development

Bad White [126]

Answer:

a) Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

b) $X \sim Bin(n =20, p=0.8)$

c) $P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

d) $P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

e) $E(X) = np = 20*0.8 = 16$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

Let the random variable X= "number of these tracks where SBIRS detects the object." in order to use the binomial probability distribution we need to satisfy some conditions:

1) Independence between the trials (satisfied)

2) A value of n fixed , for this case is 20 (satisfied)

3) Probability of success p =0.2 fixed (Satisfied)

So then we have all the conditions and we can assume that:

$X \sim Bin(n =20, p=0.8)$

Part b

$X \sim Bin(n =20, p=0.8)$

Part c

For this case we just need to replace into the mass function and we got:

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

Part d

For this case we want this probability: $P(X\geq 15)$

And we can solve this using the complement rule:

$P(X \geq 15) = P(X=15)+ .....+P(X=20)$

$P(X=15)=(20C15)(0.8)^{15} (1-0.8)^{20-15}=0.17456$

$P(X=16)=(20C16)(0.8)^{16} (1-0.8)^{20-16}=0.218$

$P(X=17)=(20C17)(0.8)^{17} (1-0.8)^{20-17}=0.205$

$P(X=18)=(20C18)(0.8)^{18} (1-0.8)^{20-18}=0.137$

$P(X=19)=(20C19)(0.8)^{19} (1-0.8)^{20-19}=0.058$

$P(X=20)=(20C20)(0.8)^{20} (1-0.8)^{20-20}=0.012$

$P(X\geq 15)=0.804208$

Part e

The expected value is given by:

$E(X) = np = 20*0.8 = 16$

5 0

3 years ago

An airplane took 5 hours to travel 3,420 kilometers.how many kilometers did the plane travel in one hour

tino4ka555 [31]

Answer:

684 Hours.

Step-by-step explanation:

If the plane travelled that distance in 5 hours, then all you need to do is divide the distance by 5 to get the distance travelled in one hour. This is obviously presuming that the speed travelled was constant or that this is an average figure.

3420/5 = 684.

8 0

3 years ago

Read 2 more answers