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4 years ago

H - 20 Solve for j in the literal equation =k.

Mathematics

1 answer:

Aloiza [94]4 years ago

6 0

h - 20/j = k solve for j

isolate each term individually $-\frac{20}{j}$ then $h$ then $-20$

$h-\frac{20}{j}=k\\-\frac{20}{j}=k-h\\j=-\frac{20}{k-h}$

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telo118 [61]

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D)6 units

Step-by-step explanation:

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Answer:

You will use the z score to find this so need a z score chart.

Step-by-step explanation: Remember to take what you want (35 ounce) minus the mean( 29) then divide by the standard deviation(4). Then you look up this number on the z score chart to find your probability of less than 35 ounces. You could then do the same thing using 34 ounces since your final answer will be between these two numbers.

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3 years ago

Drew jumped farther than all 4 students above but jumped shorter than 7 feet , 7 inches. How far could drew have jumped? Write s

S_A_V [24]

<h3>The distance covered by Drew in long jump is either 7 ft 5 inches OR 7 ft 6 inches.</h3>

Step-by-step explanation:

Here, the needed table is attached for the reference.

Now, as we can see from the table:

The distance covered by Cindy = 2 yards, 1 foot 3 inches

The distance covered by Tyrette = 7 feet, 2 inches

The distance covered by Nina = 2 yards, 1 foot, 1 inch

The distance covered by Monique = 7 feet, 4 inches

As we know : 1 yard = 3 feet

So, the distance covered by :

Cindy = 2 yards, 1 foot 3 inches = 3 ft x (2) + 1 ft + 3 in = 7 ft 3 in

Tyrette = 7 feet, 2 inches

Nina = 2 yards, 1 foot, 1 inch = = 3 ft x (2) + 1 ft + 1 in = 7 ft 1 in

Monique = 7 feet, 4 inches

So by comparing all distances, we can see that:

The maximum distances jumped by all four is 7 ft 4 inches.

The distance covered by Drew is less than 7 ft 7 in.

So, he must have jumped 7 ft 5 inches OR 7 ft 6 inches.

Hence, the distance covered by Drew in long jump is either 7 ft 5 inches OR 7 ft 6 inches.

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4 years ago

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Answer:

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3 years ago

Customers arrive at a grocery store at an average of 2.1 per minute. Assume that the number of arrivals in a minute follows the

Black_prince [1.1K]

Answer:

a) P(2)=0.270

b) P(X>3)=0.605

c) P=0.410

Step-by-step explanation:

We know that customers arrive at a grocery store at an average of 2.1 per minute. We use the Poisson distribution:

$\boxed{P(k)=\frac{\lambda^k \cdot e^{-\lambda}}{k!}}$

a) In this case: $\lambda=2.1$

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Therefore, the probability is P(2)=0.270.

b) In this case: $\lambda=2\cdot 2.1=4.2$

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Therefore, the probability is P(X>3)=0.605.

c) We know that two customers came in in the first minute. That is why we calculate the probability of at least 5 customers entering the other 2 minutes.

In this case: $\lambda=2\cdot 2.1=4.2$

$P(X\geq 5)=1-P(X$

Therefore, the probability is P=0.410.

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3 years ago