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1 year ago

Solve for n in the proportion 7/3 = n/21.

1 answer:

7 0

Given ,

7/3 = n/21

We need to find the value of n,

Cross multiplying,we get

3n = 21 x 7

=> 3n = 147

=> n = 147/3

=> n = 49 is the required answer

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Does anyone know a fraction equation that equals 13?

taurus [48]

Answer:

11 1/2 + 1 1/2 = 13

Step-by-step explanation:

11+1=12

1/2+ 1/2 =1

12+1= 13

7 0

3 years ago

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V1/V2=T1/T2. Solve for V1

Ahat [919]

V1=(T1/T2)*V2
In this problem, since you are given no actual numbers, the only thing you can do is multiply V2 over to get V1 alone. Since you can't cancel anything out, you are left with the answer above.

4 0

3 years ago

What’s the answer I really need help????

Brums [2.3K]

You should use the calculator and just plug in those numbers

5 0

3 years ago

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At winds speeds above 1000 centimeters per second (cm/sec), significant sand-moving events begin to occur. Wind speeds below 100

alisha [4.7K]

Answer:

Lower limit = 1007.95 cm/s

Upper limit = 1142.05 cm/s

Step-by-step explanation:

We have to calculate a 95% CI with population's standard deviation known.

The sample is size n=60, with mean M=1075.

The confidence interval can be expressed as:

$M-z*\sigma/\sqrt{n} \leq\mu\leq M+z*\sigma/\sqrt{n}$

For a 95% CI, the value of z=1.96.

$1075-1.96*265/\sqrt{60} \leq\mu\leq 1075+1.96*265/\sqrt{60} \\\\1075-67.05 \leq\mu\leq 1075+67.05\\\\ 1007.95 \leq\mu\leq 1142.05$

8 0

3 years ago

Newborn males have weights with a mean of 3272.8 g and a standard deviation of 660.2 g. Newborn females have weights with a mean

Maksim231197 [3]

The correct option is: a female who weighs 1500 g

Explanation

Formula for finding the z-score is: $z= \frac{X-\mu}{\sigma}$

Newborn males have weights with a mean $(\mu)$ of 3272.8 g and a standard deviation $(\sigma)$ of 660.2 g.

So, the z-score for the newborn male who weighs 1500 g will be.......

$z(X=1500)=\frac{1500-3272.8}{660.2}=-2.685... \approx -2.69$

According to the normal distribution table, $P(z=-2.69)=0.0036 = 0.36\%$

Now, newborn females have weights with a mean $(\mu)$ of 3037.1 g and a standard deviation $(\sigma)$ of 706.3 g.

So, the z-score for the newborn female who weighs 1500 g will be.......

$z(X=1500)=\frac{1500-3037.1}{706.3}=-2.176... \approx -2.18$

According to the normal distribution table, $P(z=-2.18)=0.0146 = 1.46\%$

As we can see that the probability that a newborn female has weight of 1500 g is greater than newborn male, so a newborn female has the weight of 1500 g that is more extreme relative to the group from which he came.

5 0

3 years ago