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3 years ago

Solve for t. You must write your answer in fully simplified form. -6= -12t

Mathematics

2 answers:

alexdok [17]3 years ago

6 0

T = 1/2

-6 = -12t
-6/-12 = t
6/12 = t
1/2 = t

lutik1710 [3]3 years ago

6 0

Answer:

t =1/2

Step-by-step explanation:

-6= -12t

Divide each side by -12

-6/-12 = -12t/-12

1/2 = t

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2 years ago

Are all quadrilaterals either rectangles or squares?

Greeley [361]

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3 years ago

PLEASE HELP ME RN PLEASE

Advocard [28]

Answer:

$\frac{n^{2} - 1 }{2}$

$\sqrt{29}$

Step-by-step explanation:

Area of triangle=1/2×b×h

$\frac{1}{2} \times (n - 1)(n + 1)$

$\frac{( {n}^{2} + n - n - 1) }{2 }$

$\frac{ {n}^{2} - 1 }{2}$

2. Area of triangle=14

$\frac{ {n}^{2} - 1 }{2} = 14$

${n}^{2} - 1 = 28$

${n}^{2} = 29$

$n = \sqrt{29}$

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2 years ago

ABC is a dilation image of DEF. What is the scale factor?

GrogVix [38]

The complete question in the attached figure

we know that
length side AB=8 units
length side DE=4 units

[ABC]=[DEF]*[scale factor]
then
[scale factor ]=[ABC]/[DEF]---------> 8/4--------> 2

the answer is
the scale factor for a dilation image of DEF to obtain ABC is 2

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4 years ago

A fair coin is to be tossed 20 times. Find the probability that 10 of the tosses will fall heads and 10 will fall tails, (a) usi

lbvjy [14]

Using the distributions, it is found that there is a:

a) 0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

b) 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

c) 0.1742 = 17.42% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item a:

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

20 tosses, hence $n = 20$ .
Fair coin, hence $p = 0.5$ .

The probability is P(X = 10), thus:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 10) = C_{20,10}.(0.5)^{10}.(0.5)^{10} = 0.1762$

0.1762 = 17.62% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item b:

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with $\mu = np, \sigma = \sqrt{np(1-p)}$ .

The probability of an exact value is 0, hence 0% probability that 10 of the tosses will fall heads and 10 will fall tails.

Item c:

For the approximation, the mean and the standard deviation are:

$\mu = np = 20(0.5) = 10$

$\sigma = \sqrt{np(1 - p)} = \sqrt{20(0.5)(0.5)} = \sqrt{5}$

Using continuity correction, this probability is $P(10 - 0.5 \leq X \leq 10 + 0.5) = P(9.5 \leq X \leq 10.5)$ , which is the p-value of Z when X = 10.5 subtracted by the p-value of Z when X = 9.5.