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2 years ago

How many terms are in the algebraic expression Negative 7 + 12 x 4 minus 5 y 8 + x?

Mathematics

2 answers:

BARSIC [14]2 years ago

5 0

Answer:

Therefore,

There are Four i.e 4 terms in the expression,

$-7+12x^{4}-5y^{8}+x$

Step-by-step explanation:

Algebraic Expression:

An algebraic expression is a mathematical expression that consists of variables, numbers and operations.

Variables with the coefficient is called as Term.
Number is also a term.

Here the Expression is

$-7+12x^{4}-5y^{8}+x$

So,

$-7 = One\ Term\\12x^{4}=Second\ Term\\-5y^{8}=Third\ Term\\x=Fourth\ Term$

In all there are Four i.e 4 terms in the given expressions.

Therefore,

There are Four i.e 4 terms in the expression,

$-7+12x^{4}-5y^{8}+x$

aleksandr82 [10.1K]2 years ago

4 0

4 is the answer and no explanation

hope this helps and on edgenuity it is D

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Mercedes receives a $25 gift card for downloading music and wants to determine how many songs she can purchase. Each downloaded

Contact [7]

Answer:

m less than or equal to 19

She can buy at most 19

Step-by-step explanation:

25 divided by 1.29 is 19.379844... but you can't buy fractions of a music song.

So the closest to the maximum would be 19.

6 0

1 year ago

Read 2 more answers

Orange M&M’s: The M&M’s web site says that 20% of milk chocolate M&M’s are orange. Let’s assume this is true and set

SOVA2 [1]

Answer:

The correct option is (A).

Step-by-step explanation:

Let X = number of orange milk chocolate M&M’s.

The proportion of orange milk chocolate M&M’s is, p = 0.20.

The number of candies in a small bag of milk chocolate M&M’s is, n = 55.

The event of an milk chocolate M&M being orange is independent of the other candies.

The random variable X follows a Binomial distribution with parameter n = 55 and p = 0.20.

The expected value of a Binomial random variable is:

$E(X)=np$

Compute the expected number of orange milk chocolate M&M’s in a bag of 55 candies as follows:

$E(X)=np$

$=55\times 0.20\\=11$

It is provided that in a randomly selected bag of milk chocolate M&M's there were 14 orange ones, i.e. the proportion of orange milk chocolate M&M's in a random bag was 25.5%.

This proportion is not surprising.

This is because the average number of orange milk chocolate M&M’s in a bag of 55 candies is expected to be 11. So, if a bag has 14 orange milk chocolate M&M’s it is not unusual at all.

All unusual events have a very low probability, i.e. less than 0.05.

Compute the probability of P (X ≥ 14) as follows:

$P(X\geq 14)=\sum\limits^{55}_{x=14}{{55\choose x}0.20^{x}(1-0.20)^{55-x}}$

$=0.1968$

The probability of having 14 or more orange candies in a bag of milk chocolate M&M’s is 0.1968.

This probability is quite larger than 0.05.

Thus, the correct option is (A).

4 0

2 years ago

Ana wrote four statements to describe circle S. Statements: AB is a diameter. ST , SP and SQ are radii. SQ = 12 cm PQ = 12 cm Ar

wel

Two of the four statements that Ana wrote are correct. Number 1, "AB is a diameter" is incorrect, and so is number 3, "SQ = 12 cm". AB is not a diameter because it is instead a chord. "ST , SP and SQ are radii" is correct because they are straight lines from the center of the circle to the circumference of the circle, which is the exact definition of radius. The third statemet is incorrect because since ST is a radius and it equals 6, that means all radii are equal to 6. SQ is a radius of circle S, so it should also equal 6, not 12. The last statement is correct because PQ is a diameter of circle S. By rule, the diameter is always equal to double of the radius. The radius is 6, so 6 x 2= 12.

8 0

1 year ago

Complete the equation of the line through (-6,5)(−6,5)left parenthesis, minus, 6, comma, 5, right parenthesis and (-3,-3)(−3,−3)

goldenfox [79]

Answer:

The answer to your question is: y = -8/3 x - 11

Step-by-step explanation:

A (-6, 5)

B (-3, -3)

Process

1.- Find the slope of the line

Formula

$m = \frac{y2 - y1}{x2 - x1}$

Substitution

$m = \frac{-3 - 5}{-3 + 6}$

Simplifying

$m = \frac{- 8 }{3}$

2.- Find the equation of the line

Formula

( y - y1) = m(x - x1)

Substitution

( y + 3) = -8/3 (x + 3)

Simplifying

y + 3 = -8/3 x - 8

y = -8/3 x - 8 - 3

y = -8/3 x - 11

4 0

1 year ago

It is believed that as many as 23% of adults over 50 never graduated from high school. We wish to see if this percentage is the

JulijaS [17]

Answer:

1) n=48

2) n=298

3) n=426

Step-by-step explanation:

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".

$p$ represent the real population proportion of interest

$\hat p$ represent the estimated proportion for the sample

n is the sample size required (variable of interest)

$z$ represent the critical value for the margin of error

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{\hat p(1-\hat p)}{n}})$

Part 1

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 90% of confidence, our significance level would be given by $\alpha=1-0.90=0.10$ and $\alpha/2 =0.05$ . And the critical value would be given by:

$z_{\alpha/2}=-1.64, z_{1-\alpha/2}=1.64$

The margin of error for the proportion interval is given by this formula:

$ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$ (a)

And on this case we have that $ME =\pm 0.1$ and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}$ (b)

We can assume that the estimated proportion is 0.23 for the 25 to 30 group. And replacing into equation (b) the values from part a we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.1}{1.64})^2}=47.63$

And rounded up we have that n=48

Part 2

The margin of error on this case changes to 0.04 so if we use the same formula but changing the value for ME we got:

$n=\frac{0.23(1-0.23)}{(\frac{0.04}{1.64})^2}=297.7$

And rounded up we have that n=298

Part 3

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by: