Ask question

Ask question

All categories

3 years ago

7

The women's recommended weight formula from harvard pilgrim healthcare says, "give yourself 100 lb for the first 5 ft plus 5 lb

for every inch over 5 ft tall."

1 answer:

romanna [79]3 years ago

3 0

This would have to apply for a certain height compared to a weight or vice versa. I'm 5'8'' and for me that would mean I would have a healthy weight of 140 lbs.

You might be interested in

Suppose the expected tensile strength of type-A steel is 103 ksi and the standard deviation of tensile strength is 7 ksi. For ty

ExtremeBDS [4]

Answer:

a

i So the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

ii So the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

b

the approximate distribution of $\=X - \= Y$ is $E (\= X - \= Y) = -2$ and $\sigma_{\= X - \=Y}=1.029$

Here we can see that the mean of the approximate distribution is negative which tell us that this negative value of the data for $\=X - \= Y$ sample are more and their frequency occurrence is higher than the positive values

c

the value of $P(-1 \le \=X - \= Y \le 1)$ is $= -0.1639$

Step-by-step explanation:

From the question we are given that

The expected tensile strength of the type A steel is $\mu_A = 103 ksi$

The standard deviation of type A steel is $\sigma_A = 7ksi$

The expected tensile strength of the type B steel is $\mu_B = 105\ ksi$

The standard deviation of type B steel is $\sigma_B = 5 \ ksi$

Also the assumptions are

Let $\= X$ be the sample average tensile strength of a random sample of 80 type-A specimens

Here $n_a =80$

Let $\= Y$ be the sample average tensile strength of a random sample of 60 type-B specimens.

Here $n_b = 60$

Let the sampling distribution of the mean be

$\mu _ {\= X} = \mu$

$=103$

Let the sampling distribution of the standard deviation be

$\sigma _{\= X} = \frac{\sigma }{\sqrt{n_a} }$

$= \frac{7}{\sqrt{80} }$

$=0.783$

So What this mean is that the approximate distribution of $\= X$ is $\mu_{\= X} =103$ and $\sigma_{\= X} = 0.783$

For $\= Y$

The sampling distribution of the sample mean is

$\mu_{\= Y} = \mu$

$= 105$

The sampling distribution of the standard deviation is

$\sigma _{\= Y} = \frac{\sigma }{\sqrt{n_b} }$

$= \frac{5}{\sqrt{60} }$

$= 0.645$

So What this mean is that the approximate distribution of $\= Y$ is $\mu_{\= Y} =105$ and $\sigma_{\= Y} = 0.645$

Now to obtain the approximate distribution for $\=X - \= Y$

$E (\= X - \= Y) = E (\= X) - E(\= Y)$

$= \mu_{\= X} - \mu_{\= Y}$

$= 103 -105$

$= -2$

The standard deviation of $\=X - \= Y$ is

$\sigma_{\= X - \=Y} = \sqrt{\sigma_{\= X}^2 - \sigma_{\= Y}^2}$

$= \sqrt{(0.783)^2 + (0.645)^2}$

$=1.029$

Now to find the value of $P(-1 \le \=X - \= Y \le 1)$

Let us assume that $F = \= X - \= Y$

$P(-1 \le F \le 1)$ $= P [\frac{-1 -E (F)}{\sigma_F} \le Z \le \frac{1-E(F)}{\sigma_F} ]$

$= P[\frac{-1-(-2)}{1.029} \le Z \le \frac{1-(-2)}{1.029} ]$

$= P[0.972 \le Z \le 2.95]$

$= P(Z \le 0.972) - P(Z \le 2.95)$

Using the z-table to obtain their z-score

$= 0.8345 - 0.9984$

$= -0.1639$

3 0

3 years ago

Unit 2 (12 points)

klio [65]

1+3=4

A counterexample is an a example that proves the statement false.

1+3 are not even numbers but they equal an even one, so it just proved the statement wrong.

3 0

3 years ago

How do I solve for e?<br> 9e+4=-5e+14+13e

IrinaVladis [17]

Answer: e = 10

Step-by-step explanation: 9e + 4 = -5e + 14 + 13 e

Now lets bring the variables and numbers on a different side

Remember that when the numbers go to the other side the sign changes

so 4 - 14 = -9e + 13e -5e

-10 = -e

10 = e

if plug in both equations on a graph then you will see that when e = 10 the lines meet at 94.

7 0

3 years ago

Read 2 more answers

-11 - 5x = 6(5x + 4)

gavmur [86]

Answer:

x=-1

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

find the present value that will grow to $7000 if the annual interest rate is 3.5% compounded quarterly for 9 uears

Gwar [14]

PV=FV/(1+i)^t

FV=7000, i=3.5%=0.035,t=9

put everything in the formula
PV=7000/(1+0.035)^9
PV=$5136.12

7 0

3 years ago