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3 years ago

Help plz I really need to finish

Mathematics

1 answer:

laiz [17]3 years ago

5 0

Answer:

$26x^6y^4$

Step-by-step explanation:

the volume is like

$V = length\cdot width\cdot depth$

so we have

$V= 4yx^5\cdot 3xy\cdot 2y^2\\\\V=(4\cdot 3\cdot 2)(y\cdot y \cdot y^2)(x^5 \cdot x)\\\\V = 24(y^4)(x^6)\\\\V=24x^6y^4$

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Kaia's mom is having triplets. Each baby could be either male or female

fiasKO [112]

Answer:

MMM

Step-by-step explanation:

Given

See attachment for elements of the sample space (in a probability tree)

Required

Which of MMM, FFM, MFF, FMF represents more than one male?

From the question, we understand that:

$M \to Male$

$F \to Female$

So, for an event to represent more than 1 male, there must be an occurrence of at least 2 M (i.e 2 M or 3 M)

From the given options, only MMM has more than 1 M.

Hence, MMM represents more than one Male

5 0

3 years ago

Answerrr pleaseeeeeeeeeeeee

erastova [34]

There were 7 people

3 0

3 years ago

Leah gets paid $450 a week +5% commission on her sales she hopes to earn at least $600 each week right in any quality to represe

ElenaW [278]

x = sales

450 + 0.05x = 600

5 0

3 years ago

You were told that the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distrib

uysha [10]

Answer:

Step-by-step explanation:

Given that that (X) the amount of time lapsed between consecutive trades on the New York Stock Exchange followed a normal distribution with a mean of 15 seconds.

i.e. X is normal with mean = 15 and unknown std deviation $\sigma$

Given that $P(X$

i.e. P( $z$

z=-1.475 (from normal table)

Hence $\frac{13-15}{\sigma}=-1.475\\\sigma = 1.356$

Using this we find P(X>17) = $P(Z>\frac{17-15}{1.356} \\=P(Z>1.475)\\=0.5-0.428\\=0.072$

6 0

3 years ago

Consumer products are required by law to contain at least as much as the amount printed on the package. For example, a bag of po

Troyanec [42]

Answer:

The average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802

Step-by-step explanation:

We are given that

Standard deviation, $\sigma=0.2$ ounces

We have to find the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag.

$P(x\geq 10)=0.99$

Assume the bag weight distribution is bell-shaped

Therefore,

$P(\frac{x-\mu}{\sigma}\geq 10)=0.99$

We know that

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

Using the value of z

Now,

$\frac{10-\mu}{0.2}=0.99$

$10-\mu=0.99\times 0.2$

$\mu=10-0.99\times 0.2$

$\mu=9.802$

Hence, the average bag weight must be used to achieve at least 99 percent of the bags having 10 or more ounces in the bag=9.802

5 0

3 years ago

Help plz I really need to finish​

Help plz I really need to finish