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3 years ago

14

Pls help no explanation needed

2 answers:

liq [111]3 years ago

8 0

It’s 14 ( positive) have a good day!

tensa zangetsu [6.8K]3 years ago

5 0

Answer: 70

250/50 = 5

14 x 5 = 70

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1. If B is between A and C, AB = 12 in., and AC = 19 in., then what is BC?

Oksanka [162]

Answer:

BC = 7.

Step-by-step explanation:

subtract 12 inches from 19 and then you get the result for BC.

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4 years ago

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3 years ago

Normal Distribution. Cherry trees in a certain orchard have heights that are normally distributed with mu = 112 inches and sigma

Lubov Fominskaja [6]

Answer:

The probability that a randomly chosen tree is greater than 140 inches is 0.0228.

Step-by-step explanation:

Given : Cherry trees in a certain orchard have heights that are normally distributed with $\mu = 112$ inches and $\sigma = 14$ inches.

To find : What is the probability that a randomly chosen tree is greater than 140 inches?

Solution :

Mean - $\mu = 112$ inches

Standard deviation - $\sigma = 14$ inches

The z-score formula is given by, $Z=\frac{x-\mu}{\sigma}$

Now,

$P(X>140)=P(\frac{x-\mu}{\sigma}>\frac{140-\mu}{\sigma})$

$P(X>140)=P(Z>\frac{140-112}{14})$

$P(X>140)=P(Z>\frac{28}{14})$

$P(X>140)=P(Z>2)$

$P(X>140)=1-P(Z$

The Z-score value we get is from the Z-table,

$P(X>140)=1-0.9772$

$P(X>140)=0.0228$

Therefore, the probability that a randomly chosen tree is greater than 140 inches is 0.0228.

5 0

3 years ago

For the given term, find the binomial raised to the power, whose expansion it came from: 15(5)^2 (-1/2 x) ^4

Elina [12.6K]

Answer:

<em>C.</em> $(5-\frac{1}{2})^6$

Step-by-step explanation:

Given

$15(5)^2(-\frac{1}{2})^4$

Required

Determine which binomial expansion it came from

The first step is to add the powers of he expression in brackets;

$Sum = 2 + 4$

$Sum = 6$

Each term of a binomial expansion are always of the form:

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

Where n = the sum above

$n = 6$

Compare $15(5)^2(-\frac{1}{2})^4$ to the above general form of binomial expansion

$(a+b)^n = ......+15(5)^2(-\frac{1}{2})^4+.......$

Substitute 6 for n

$(a+b)^6 = ......+15(5)^2(-\frac{1}{2})^4+.......$

[Next is to solve for a and b]

<em>From the above expression, the power of (5) is 2</em>

<em>Express 2 as 6 - 4</em>

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

By direct comparison of

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

and

$(a+b)^6 = ......+15(5)^{6-4}(-\frac{1}{2})^4+.......$

We have;

$^nC_ra^{n-r}b^r= 15(5)^{6-4}(-\frac{1}{2})^4$

Further comparison gives

$^nC_r = 15$

$a^{n-r} =(5)^{6-4}$

$b^r= (-\frac{1}{2})^4$

[Solving for a]

By direct comparison of $a^{n-r} =(5)^{6-4}$

$a = 5$

$n = 6$

$r = 4$

[Solving for b]

By direct comparison of $b^r= (-\frac{1}{2})^4$

$r = 4$

$b = \frac{-1}{2}$

Substitute values for a, b, n and r in

$(a+b)^n = ......+ ^nC_ra^{n-r}b^r+.......$

$(5+\frac{-1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ ^6C_4(5)^{6-4}(\frac{-1}{2})^4+.......$

Solve for $^6C_4$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{(6-4)!4!)}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6!}{2!!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5*4!}{2*1*!4!}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{6*5}{2*1}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+ \frac{30}{2}*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15*(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^{6-4}(\frac{-1}{2})^4+.......$

$(5-\frac{1}{2})^6 = ......+15(5)^2(\frac{-1}{2})^4+.......$

<em>Check the list of options for the expression on the left hand side</em>

<em>The correct answer is </em> $(5-\frac{1}{2})^6$ <em />

3 0

3 years ago

Please help me with this question

Lady_Fox [76]

Answer:

8 = 2^3, 52 = 2^2 * 13

Step-by-step explanation:

8 = 2 * 2 * 2 = 2^3

52 = 2 * 2 * 13= 2^2 * 13

4 0

2 years ago

Read 2 more answers