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3 years ago

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A small box of chocolates weighs a fifth a large box of chocolates. The difference in weight between a small box of chocolate an

d large box of chocolates is 200 grams. Find the weight of the large box of chocolate

1 answer:

antoniya [11.8K]3 years ago

7 0

Answer:

4 punds

Step-by-step explanation:

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If f(x)=2x-5, which expression represents f(x+1)? A) 2x-3, B) 2x-4, C)2x-5, D)2x+7

Dovator [93]

F(x)=x^2+2x+1 & g(x)=3(x+1)^2
now, f(x)+g(x)
=x^2+2x+1+3(x+1)^2
=x^2+2x+1+3(x^2+2x+1)
=x^2+2x+1+3x^2+6x+3
=4x^2+8x+4<===answer(c)
next:
f(x)=x^2-1 & g(x)=x+3
now, f(g(x))=(x+3)^ -1
=x^2+6x+9-1
=x^2+6x+8<====answer(b)
i solve two of ur problems.
now try the 3rd one that is similar to no. 1
and try the last two urself.

5 0

3 years ago

90 points please help!!!!!!!!

just olya [345]

To model this situation, we are going to use the exponential function: $f(t)=P(1+ \frac{r}{n} )^{nt}$
where
$P$ is the initial number of cars
$r$ is the growing rate in decimal form
$n$ is number of tames the growing rate is increasing per year
$t$ is the time in years

To convert the growing rate to decimal form, we are going to divide the rate by 100%
$r= \frac{12}{100}$
$r=0.12$
Since the growing rate is increasing quarterly, $n=4$ . We also know that the initial number of cars is 920, so $P=920$ . Lets replace all those values in our function:
$f(t)=920(1+ \frac{0.12}{4} )^{4t}$
$f(t)=920(1+0.03)^{4t}$
$f(t)=900(1.03)^{4t}$

We can conclude that:
Rate ---------> The quarterly rate of growth is 0.03 or 3%
Exponent --------> The compound periods multiplied by the number of years is 4t
Coefficient--------> The initial number of cars serviced is 920
Base------> The growth factor is represented by 1.03

3 0

3 years ago

Read 2 more answers

If 10+9=21 than is 1+1= 2?

Andre45 [30]

I guess so, no wait no it’s not jk lol

4 0

3 years ago

If x=12 What is the value of the expression2(x)+3

GaryK [48]

Answer:

147

Step-by-step explanation:

2(X)= (X) x (X)

12 x 12 + 3= 144+3=147

7 0

3 years ago

PLEASE HELP ME I BEG I WILL GIVE BRAINLIEST

ohaa [14]

Answer:

Mean = (2.2 + 2.4 + 2.5 + 2.5 + 2.6 + 2.7)/6 = 2.48

Standard deviation = √(summation(x - mean)²/n

n = 6

Summation(x - mean)² = (2.2 - 2.48)^2 + (2.4 - 2.48)^2 + (2.5 - 2.48)^2 + (2.5 - 2.48)^2 + (2.6 - 2.48)^2 + (2.7 - 2.48)^2 = 0.1484

Standard deviation = √(0.1484/6

s = 0.16

Standard error = s/√n = 0.16/√6 = 0.065

Part B

Confidence interval is written as sample mean ± margin of error

Margin of error = z × s/√n

Since sample size is small and population standard deviation is unknown, z for 98% confidence level would be the t score from the student t distribution table. Degree of freedom = n - 1 = 6 - 1 = 5

Therefore, z = 3.365

Margin of error = 3.365 × 0.16/√6 = 0.22

Confidence interval is 2.48 ± 0.22

Part C

We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

H0: µ = 2.3

For the alternative hypothesis,

H1: µ > 2.3

This is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 6

Degrees of freedom, df = n - 1 = 6 - 1 = 5

t = (x - µ)/(s/√n)

Where

x = sample mean = 2.48

µ = population mean = 2.3

s = samples standard deviation = 0.16

t = (2.48 - 2.3)/(0.16/√6) = 2.76

We would determine the p value using the t test calculator. It becomes

p = 0.02

Assuming significance level, alpha = 0.05.

Since alpha, 0.05 > than the p value, 0.02, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the sample data showed significant evidence that the mean absolute refractory period for all mice when subjected to the same treatment increased.

Step-by-step explanation:

7 0

3 years ago