Simplify the expression. 10P8

A study analyzed the average yearly salt intake in the United States from 2012 thru 2017. The data is summarized in the table: Y

Nuetrik [128]

Answer:

Is an estimate of the average grams of salt used in 2012

Step-by-step explanation:

Let

x ---> the number of years since 2012

f(x) ---> is the total amount of salt ingested in grams

we have

$f(x)=52.3x+3.548$

This is a linear equation in slope intercept form

where

The slope is equal to

$m=52.3\ grams/ year$

The y-intercept or initial value is equal to

$b=3,548\ grams$

The y-intercept is the value of the function f(x) when the value of x is equal to zero

In this context, the y-intercept is the average amount of grams of salt ingested in the year 2012

The exact value of the average amount of grams of salt ingested in the year 2012 is 3,554 grams (see the data in the table)

Compare the exact value with the y-intercept

$3,554\ g\neq 3,548\ g$

therefore

3,548 grams Is an estimate of the average grams of salt used in 2012

7 0

3 years ago

That sounds like a good deal, but how about I give you 10 points for an answer.

Ilia_Sergeevich [38]

The answer is A - 1
Hope it helped

3 0

2 years ago

Read 2 more answers

EASY— get marked brainliest!!!

Ilya [14]

Answer:

y≥1/2 x +1

Step-by-step explanation:

first the line in the graph is bold and not dotted

so the sign is either ≤ or≥

the shaded area is on the top , on this case it is greater than

the right answer is y≥1/2 x +1

7 0

3 years ago

Which shows the list of numbers in order from least to greatest?

Minchanka [31]

Answer:

The answer is A.-2, |-4/5|, |-1|, |3.5|, |-4.2|

Step-by-step explanation:

3 0

2 years ago

Of a group of randomly selected adults, 360 identified themselves as manual laborers, 280 identified themselves as non-manual wa

Wittaler [7]

Answer:

We are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

Step-by-step explanation:

We are given that in a group of randomly selected adults, 160 identified themselves as executives.

n = 160

Also we are given that 42 of executives preferred trucks.

So the proportion of executives who prefer trucks is given by

p = 42/160

p = 0.2625

We are asked to find the 95% confidence interval for the percent of executives who prefer trucks.

We can use normal distribution for this problem if the following conditions are satisfied.

n×p ≥ 10

160×0.2625 ≥ 10

42 ≥ 10 (satisfied)

n×(1 - p) ≥ 10

160×(1 - 0.2625) ≥ 10

118 ≥ 10 (satisfied)

The required confidence interval is given by

$p \pm z\times \sqrt{\frac{p(1-p)}{n} }$

Where p is the proportion of executives who prefer trucks, n is the number of executives and z is the z-score corresponding to the confidence level of 95%.

Form the z-table, the z-score corresponding to the confidence level of 95% is 1.96

$p \pm z\times \sqrt{\frac{p(1-p)}{n} }$

$0.2625 \pm 1.96\times \sqrt{\frac{0.2625(1-0.2625)}{160} }$

$0.2625 \pm 1.96\times 0.03478$

$0.2625 \pm 0.06816$

$0.2625 - 0.06816, \: 0.2625 + 0.06816$

$(0.1943, \: 0.3306)$

$(19.43\%, \: 33.06\%)$

Therefore, we are 95% confident that the percent of executives who prefer trucks is between 19.43% and 33.06%

5 0

3 years ago