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3 years ago

Make k the subject of 7k=4k/3t-11t

Mathematics

2 answers:

dimaraw [331]3 years ago

6 0

Multiply 4k to 7k. Then divide 3t-11t by 7x4. Then square root the rest

love history [14]3 years ago

3 0

We are given the equation $7k=\frac{4k}{3t}-11t$

In order to make k the subject, we will have to isolate k and express it in terms of the other variable, t. Let us begin:

Multiply both sides by 3t

$21kt=4k-33t^2$

Let us now subtract 4k from both sides:

$21kt-4k=-33t^2$

Take k out as the common factor on the left hand side of the above equation:

$k(21t-4)=-33t^2$

Divide both sides by 21t-4

$k=\frac{-33t^2}{21t-4}$

If we multiply the numerator and the denominator on the right side by -1 we will get, k to be as:

$k=\frac{33t^2}{4-21t}$

The above the given equation expressed with k as the subject.

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3 years ago

Which represents where f(x) = g(x)?

Lyrx [107]

The correct answer is:

f(2) = g(2) and f(0) = g(0)

Explanation:

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4 years ago

Read 2 more answers

Eva decorated cupcakes with pound of red jelly beans and pound of blue jelly beans. She says that she used pound of jelly beans

nikklg [1K]

Answer:

No, her statement is not reasonable. This is because, she refused to acknowledge the fact that she used a pound of Blue Jelly during the decoration of the cake.

Skipping this information makes is dangerous for others who might not be aware of the Blue Jelly and eat the cupcakes thereby causing them problems (Maybe, they are allergic to Blue Jelly)

Step-by-step explanation:

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3 years ago

Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution. At Meadowbrook Hospital, the mean we

Marina86 [1]

Answer:

Required Probability = 0.1283 .

Step-by-step explanation:

We are given that at Meadow brook Hospital, the mean weight of babies born to full-term pregnancies is 7 lbs with a standard deviation of 14 oz.

Firstly, standard deviation in lbs = 14 ÷ 16 = 0.875 lbs.

Also, Birth weights of babies born to full-term pregnancies follow roughly a Normal distribution.

Let X = mean weight of the babies, so X ~ N( $\mu = 7 lbs , \sigma^{2} = 0.875^{2} lbs$ )

The standard normal z distribution is given by;

Z = $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, X bar = sample mean weight

n = sample size = 4

Now, probability that the average weight of the four babies will be more than 7.5 lbs = P(X bar > 7.5 lbs)

P(X bar > 7.5) = P( $\frac{Xbar-\mu}{\frac{\sigma}{\sqrt{n} } }$ > $\frac{7.5-7}{\frac{0.875}{\sqrt{4} } }$ ) = P(Z > 1.1428) = 0.1283 (using z% table)

Therefore, the probability that the average weight of the four babies will be more than 7.5 lbs is 0.1283 .

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3 years ago

The function f(x) = x2 – 6x + 9 is shifted 5 units to the left to create g(x). what is g(x)?

allsm [11]

The correct answer is C. The form for shifts is (x-h)^2 +k where h=the x variable ( horizontal shift left or right) and k = the y variable (vertical shift up or down) The confusing part for many students is remember that if you are moving left it's subtraction So in the form x-h it is actually x - (-h) so it gets turned into a positive/ addition sign.

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3 years ago