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3 years ago

A student writes 5y*3 to model the relationship the sum of 5y and 3. explain the error

2 answers:

5 0

5y*3
This models the relationship as 5y times 3

But the question says to model the relationship as 5y plus 3.

Therefore, the right answer is 5y+3

!!!HOPE THIS HELPS!!!

mezya [45]3 years ago

4 0

So,

5y*3 is the open phrase the student uses to model "the sum of 5y and 3".

"The sum of" means addition. The student put 5y*3, while the sum of 5y and 3 is actually 5y + 3.

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3 years ago

Situation:

Gala2k [10]

Answer:

5.5 days (nearest tenth)

Step-by-step explanation:

Given formula:

$\sf N=N_0e^{-kt}$

$\sf N_0$ = initial mass (at time t=0)
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k = a positive constant
t = time (in days)

Given values:

$\sf N_0$ = 11 g
k = 0.125

Half-life: The time required for a quantity to reduce to half of its initial value.

To find the time it takes (in days) for the substance to reduce to half of its initial value, substitute the given values into the formula and set N to half of the initial mass, then solve for t:

$\begin{aligned}\sf N & = \sf N_0e^{-kt}\\\\\implies \sf \dfrac{11}{2} & = \sf 11e^{-0.125t}\\\\\sf \dfrac{1}{2} & = \sf e^{-0.125t}\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf \ln e^{-0.125t\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t \ln e\\\\\sf \ln \left(\dfrac{1}{2}\right) & = \sf -0.125t(1)\\\sf t & = \sf \dfrac{\ln \left(\dfrac{1}{2}\right)}{-0.125}\\\\\sf t & = \sf 5.545177444...\\\\\sf t & = \sf 5.5\:\:days\:\:(nearest\:tenth)\end{aligned}$

Therefore, the substance's half-life is 5.5 days (nearest tenth).

Learn more about solving exponential equations here:

brainly.com/question/28016999

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2 years ago

Read 2 more answers

You need to construct an open-top rectangular box with a square base that must hold a volume of exactly 475 cm3. The material fo

zepelin [54]

Answer:

The dimensions of the box are:

x = 8,93 cm and h = 5,95 cm

C(min) = 850,69 cents

Step-by-step explanation:

The volume of the box is:

V = x²*h where x is the side of the square base and h the height

then h = V/ x² ⇒ h = 475 / x²

The total cost of box C is:

C = C₁ + 4*C₂ Where C₁ and C₂ are the costs of the base and one lateral side respectevily

Then cost C = 8*x² + 4* 6*h*x

The cost C as a function of x is

C(x) = 8*x² + (24* 475 /x² )*x

C(x) = 8*x² + 11400/x

Tacking derivatives on both sides of the equation

C´(x) = 16*x - 11400/x²

C´(x) = 0 ⇒ 16*x - 11400/x² = 0

16*x³ = 11400 ⇒ x³ = 11400/16

x³ = 712,5

x = 8,93 cm

and h = 475 / (8,93)² ⇒ h = 5,95 cm

C(min) = 8*79,77 + 4* ( 8,93)*5,95

C(min) = 638,16 + 212,53

C(min) = 850,69 cents

To check if value x = 8,93 would make C(x) minimum we go to the second derivatives

C´´(x) = 16 + 22800/x³ > 0

Then we have a minimum of C at x = 8,93

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3 years ago

Two companies, A and B, make express delivery for small-item packages in a city. Company A charges a flat fee of RM70 per packag

bazaltina [42]

Using the normal probability distribution and the central limit theorem, it is found that:

a) There is a 0.1922 = 19.22% probability that Company B would charge more than Company A to deliver a small-item package.

b) The mean amount charged is of RM 51.5, with a standard deviation of 21.35.

In a normal distribution with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

It measures how many standard deviations the measure is from the mean.
After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.

By the Central Limit Theorem:

When a fixed constant k multiplies a variable, the mean is $k\mu$ and the standard deviation is $k\sigma$
When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.