D. 12:28

Mathematics

1 answer:

Luda [366]2 years ago

7 0

1.25 cups per serving

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Two cylinders are similar. The radius of cylinder A is 5.6 inches. The radius of cylinder B is 1.4 inches. If the height of cyli

natita [175]

The answer is 16 inches.

Let r be a radius and h be a height of a cylinder
Since the cylinders are similar, then the ratio between their radiuses is equal to the ratio of their heights:
r1 : r2 = h1 : h2

Cylinder A:
r1 = 5.6 in
h1 = ?

Cylinder B:
r2 = 1.4 in
h2 = 4 in

5.6 : 1.4 = h1 : 4
h1 = 4 * 5.6 : 1.4
h1 = 16 inches

7 0

3 years ago

Read 2 more answers

At a fixed operating setting, the pressure in a line downstream of a reciprocating compressor has a mean value of 950 kPa with a

prisoha [69]

Answer:

4.75% probability that the line pressure will exceed 1000 kPa during any measurement

Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 950, \sigma = 30$

What is the probability that the line pressure will exceed 1000 kPa during any measurement

This is 1 subtracted by the pvalue of Z when X = 1000. So

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

$Z = \frac{1000 - 950}{30}$

$Z = 1.67$

$Z = 1.67$ has a pvalue of 0.9525

1 - 0.9525 = 0.0475

4.75% probability that the line pressure will exceed 1000 kPa during any measurement

7 0

3 years ago

Solve the equation 2 sec² x = 3 - tan x for the domain 0° ≤ x ≤ 360°

vodka [1.7K]

$2 \sec^2 x = 3- \tan x \\\\\implies 2(1 +\tan^2 x) = 3- \tan x \\\\\implies 2 + 2 \tan^2 x +\tan x -3 =0\\\\\implies 2 \tan^2 x + \tan x -1 =0\\\\\implies 2u^2 + u -1 =0~~~ ;[\text{set} \tan x = u]\\\\\implies u = \dfrac{-1\pm \sqrt{1-4\cdot 2 \cdot (-1)}}{2(2)}\\\\\implies u =\dfrac{-1 \pm\sqrt{9}}{4}\\\\\implies u = \dfrac{-1 \pm 3}{4}\\\\\implies u = \dfrac{2}{4}=\dfrac 12~~ \text{or} ~~u =\dfrac{-4}{4} =-1\\\\$

$\implies \tan x = \dfrac 12 ~~ \text{or} ~~ \tan x =-1~~~ ;[\text{Substitute back u =tan x}]\\\\\text{Now,}~ \\\\\tan x = -1,\\\\\implies x = n\pi - \dfrac{\pi}4\\\\\\\text{For interval,}~ [0,2\pi) \\\\x = \dfrac{3\pi}4, \dfrac{7\pi}4\\\\\text{In degrees,}~ x = 135^{\circ}, x =315^{\circ}\\\\\tan x = \dfrac 12\\\\\implies x = n\pi + \tan^{-1} \left(\dfrac 12 \right)\\\\\text{For interval} ~[0,2\pi),\\\\$

$x=\tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\text{in degrees,} ~~x=\tan^{-1} \left(\dfrac 12 \right), \left[180^{\circ} +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\\\\text{Combine all solutions:}\\\\x=\dfrac{3\pi}4, \dfrac{7\pi}4, \tan^{-1} \left(\dfrac 12 \right), \left[\pi +\tan^{-1} \left( \dfrac 12 \right)\right]\\\\\\$