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photoshop1234 [79]

3 years ago

8

tems produced by a manufacturing process are supposed to weigh 90 grams. However, there is variability in the items produced, an

d they do not all weigh exactly 90 grams. The distribution of weights can be approximated by a Normal distribution with a mean of 90 grams and a standard deviation of 1 gram. What percentage of the items will either weigh less than 87 grams or more than 93 grams? Group of answer choices

1 answer:

leva [86]3 years ago

5 0

Answer:

Step-by-step explanation:What percentage of organs weighs less than 285 grams or more than 315 grams?

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I need help! Is this correct

riadik2000 [5.3K]

Answer:

Yes

Step-by-step explanation:

5 0

3 years ago

Evaluate: 16.1 ÷ 4.6 - 1.78

Lana71 [14]

16.1 divided by 4.6 - 1.78

equals 1.72

Hope this helps,
QueenBeauty666

3 0

3 years ago

Read 2 more answers

A curve passes through the point (0, 5) and has the property that the slope of the curve at every point P is twice the y-coordin

Anit [1.1K]

Answer:

$y=5e^{2x}$

Step-by-step explanation:

Let (x,y) represents a point P on the curve,

So, the slope of the curve at point P = $\frac{dy}{dx}$

According to the question,

$\frac{dy}{dx}=2y$

$\frac{1}{y}dy=2dx$

Integrating both sides,

$\int \frac{dy}{y}=2dx$

$ln y=2x+ln C$

$ln y-ln C = 2x$

$ln(\frac{y}{C})=2x$

$\frac{y}{C}=e^{2x}$

$\implies y=Ce^{2x}$

Since, the curve is passing through the point (0, 5),

$5=Ce^{0}\implies C=5$

Hence, the required equation of the curve is,

$y=5e^{2x}$

7 0

3 years ago

a 10-kg object traveling at 6 m/s collides head on with a 2-kg object at rest. What is the total momentum of the system after th

Neko [114]

Answer:

M1V1i+M2V2i=(M1 + M2)vf

(10)(6)+(2)(0)=(10+2)vf

60=12vf

60÷12 =12vf÷12

5=vf

::Vf=5ms-1 to the right

7 0

3 years ago

Suppose the force acting on a column that helps to support a building is a normally distributed random variable X with mean valu

Sauron [17]

Answer:

(1) 0.4207

(2) 0.7799

Step-by-step explanation:

Given,

Mean value,

$\mu = 15.0$

Standard deviation,

$\sigma = 1.25$

(1) P(X ≥ 17.5) = 1 - P( X ≤ 17.5)

$= 1- P(\frac{x-\mu}{\sigma} \leq \frac{17.5-\mu}{\sigma})$

$=1-P(z\leq \frac{17.5 - 15}{1.25})$

$=1-P(z\leq \frac{2.5}{1.25})$

$=1-P(z\leq 2)$

$=1- 0.5793$ ( By using z-score table )

= 0.4207

(2) P(14 ≤ X ≤ 18) = P(X ≤ 18) - P(X ≤ 14)

$=P(z\leq \frac{18 - 15}{1.25}) - P(z\leq \frac{14 - 15}{1.25})$

$=P(z\leq \frac{3}{1.25}) - P(z\leq -\frac{1}{1.25})$

$=P(z\leq 2.4) - P(z\leq -0.8)$

= 0.9918 - 0.2119

= 0.7799

8 0

3 years ago