All categories

3 years ago

Explain how you can round 25.691 to the greatest place.

Mathematics

2 answers:

nadezda [96]3 years ago

6 0

Answer:

Round to nearest tens place; 30.

Step-by-step explanation:

We are asked to explain how we can round 25.691 to the greatest place.

We can see that greatest digit of our given number is tens digit that is 2.

To round our given number to greatest place, we need to round our answer to nearest tens.

Since the one digit is 5, so we will round up our answer to nearest tenth.

Therefore, our number rounded to greatest place would be $30$ .

labwork [276]3 years ago

4 0

If it is to the tenth, then you have to look at the hundredth place, if the number in the hundredth place is over 5 or 5, then you have to make the number in the tenth place move up one digit if it isn't over 5 or 5, then you don't do anything

You might be interested in

I need help plz hurry fast!!

Svetlanka [38]

Answer:

Whats the question

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Find the coordinates of P ' if D o, 2.4 (x,y) is applied to the triangle below

guajiro [1.7K]

Answer:

The coordinates of P' are (4.8,-4.8).

Step-by-step explanation:

The rule of dilation $D_{o,2.4}(x,y)$ represent the dilation with scale factor 2.4 and center at origin.

If the scale factor of the dilation is k and the center is (0,0), then

$(x,y)\rightarrow (kx,ky)$

Since the scale factor is 2.4, therefore

$(x,y)\rightarrow (2.4x,2.4y)$

From the given figure it is noticed that the coordinates of P are (2,-2). The coordinates of P' are

$P(2,-2)\rightarrow P'(2.4(2),2.4(-2))$

$P(2,-2)\rightarrow P'(4.8,-4.8)$

Therefore the coordinates of P' are (4.8,-4.8).

3 0

3 years ago

Four lawn sprinkler heads are fed by a 1.9-cm-diameter pipe. The water comes out of the heads at an angle of 35° above the horiz

klemol [59]

Answer:

Step-by-step explanation:

Given:

Angle, θ = 35°

Vertical distance, Δx = 6 m

Diameter, d = 1.9 cm

= 0.019 m

When the water leaves the sprinkler, it does so at a projectile motion.

Therefore,

Using equation of motion,

(t × Vox) = 2Vo²(sin θ × cos θ)/g

= Δx = 2Vo²(sin 35 × cos 35)/g

Vo² = (6 × 9.8)/(2 × sin 35 × cos 35)

= 62.57

Vo = 7.91 m/s

Area of sprinkler, As = πD²/4

Diameter, D = 3 × 10^-3 m

As = π × (3 × 10^-3)²/4

= 7.069 × 10^-6 m²

V_ = volume rate of the sprinkler

= area, As × velocity, Vo

= (7.069 × 10^-6) × 7.91

= 5.59 × 10^-5 m³/s

Remember,

1 m³ = 1000 liters

= 5.59 × 10^-5 m³/s × 1000 liters/1 m³

= 5.59 × 10^-2 liters/s

= 0.0559 liters/s.

For the 4 sprinklers,

The rate at which volume is flowing in the 4 sprinklers = 4 × 0.0559

= 0.224 liters/s

Area of 1.9 cm pipe, Ap = πD²/4

= π × (0.019)²/4

= 2.84 × 10^-4 m²

Volumetric flowrate of the four sprinklers = 4 × 5.59 × 10^-5 m³/s

= 2.24 × 10^-4 m³/s

Velocity of the water, Vw = volumetric flowrate/area

= 2.24 × 10^-4/2.84 × 10^-4

= 0.787 m/s

7 0

3 years ago

Read 2 more answers

The scores on the GMAT entrance exam at an MBA program in the Central Valley of California are normally distributed with a mean

Kaylis [27]

Answer:

58.32% probability that a randomly selected application will report a GMAT score of less than 600

93.51% probability that a sample of 50 randomly selected applications will report an average GMAT score of less than 600

98.38% probability that a sample of 100 randomly selected applications will report an average GMAT score of less than 600

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

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

The Z-score 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. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .