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3 years ago

Which is lowest; Brand A: 240 mg sodium for 1/3 pickle or Brand B: 325 mg sodium for 1/2 pickle.

1 answer:

8 0

Brand A is. Just divide 100 (100% of a pickle) by each number, so 100 divided by 240 and 100 by 325. Then you multiply them by the fraction of 100, which 1/3 is 33.3% and 1/2 is 50%. It comes very close but A is the lowest.

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1652÷24 is this right or wrong?

Mandarinka [93]

Answer: Yup. This is right. 68 with a remainder of 20.

Step-by-step explanation:

Nice long division and nice hand writing :)

7 0

3 years ago

Read 2 more answers

How do I solve this ? (5+i)(4-2i)

denpristay [2]

(5 + i)(4 - 2i)
20 - 10i + 4i - 2i²
20 - 6i - 2(-1)
20 - 6i + 2
20 + 2 - 6i
22 - 6i

3 0

3 years ago

A telemarketing company had a loss of $2113.15 in July, a loss of $597.11 in August, and a profit of

bixtya [17]

Yes , -2113.15 + (-597.11) + 4121.55 = 1411.29 so the total profit was 1411.29

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3 years ago

A rain drop hitting a lake makes a circular ripple. Suppose the radius, in inches, grows as a function of time in minutes accord

kolbaska11 [484]

Answer:

A(t) = 676π(t+1)

Correct question:

A rain drop hitting a lake makes a circular ripple. Suppose the radius, in inches, grows as a function of time in minutes according to r(t)=26√(t+1), and answer the following questions. Find a function, A(t), for the area of the ripple as a function of time.

Step-by-step explanation:

The area of a circle is expressed as;

A = πr^2

Where, A = Area

r = radius

From the case above.

The radius of the ripple is a function of time

r = r(t) = 26√(t+1)

So,

A(t) = π[r(t)]^2

Substituting r(t),

A(t) = π(26√(t+1))^2

A(t) = π(676(t+1))

A(t) = 676π(t+1)

3 0

3 years ago

The water content of a soil volume is measured four times gravimetrically by oven drying. The mean value of the water content is

expeople1 [14]

Answer:

99% confidence interval for the soil water content is determined by

(0.19474, 0.26926)

Step-by-step explanation:

Explanation:-

Given mean of the sample = 23.2% = 0.232

Given standard deviation of the sample = 1.0% =0.01

Given sample size 'n' =4

Degrees of freedom = n-1 =4-1 =3

critical value = 0.01

$t_{\frac{0.01}{2} } = t_{0.005} =7.4534$

99% confidence interval for the soil water content is determined by

$((x^{-} - t_{0.005} \frac{S.D}{\sqrt{n} } , (x^{-} + t_{0.005} \frac{S.D}{\sqrt{n} })$

$((0.232 - 7.4534\frac{0.01}{\sqrt{4} } , (0.232 + 7.4534 \frac{0.01}{\sqrt{4} })$

( 0.232 -0.03726 , 0.232 +0.03726)

(0.19474, 0.26926)

Final answer:-

99% confidence interval for the soil water content is determined by

(0.19474, 0.26926)

3 0

3 years ago