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2 years ago

If a cheeseburger weighs a half pound on Earth, what will it weigh on Jupiter?

Mathematics

2 answers:

beks73 [17]2 years ago

7 0

If a cheeseburger weighs half a pound on earth, it will be about 1.265 pounds on Jupiter

Katena32 [7]2 years ago

3 0

Answer:

47 pounds i think \/'_'\/

Step-by-step explanation:

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The number of bacteria in a petri dish on the first day was 113 cells. If the number of bacteria increase at a rate of 82% per d

Tresset [83]

Answer:

4107 cells

Step-by-step explanation:

From the question, we have the following values:

Day 1 : 113 cells

Number of cells increases by day by 82%

Hence,

Day 2

113 × 82% = 92.66cells

Hence, Total number of bacteria cells for Day 2 = 113 + 92.66 = 205.66cells

Day 3

205.66 × 82% = 168.6412 cells

Hence, Total number of bacteria cells for Day 3 = 168.6412 + 205.66 = 374.3012 cells

Day 4

374.3012 × 82% = 306.926984 cells

Hence, Total number of bacteria cells for Day 4 = 306.926984 + 374.3012 = 681.228184 cells

Day 5

681.228184 × 82% = 558.60711088 cells

Hence, Total number of bacteria cells for Day 5 = 558.60711088 + 681.228184 = 1239.8352949 cells

Day 6

1239.8352949 × 82% = 1016.6649418 cells

Hence, Total number of bacteria cells for Day 5 = 1016.6649418 + 1239.8352949 = 2256.5002367 cells

Day 7

2256.5002367 × 82% = 1850.3301941 cells

Hence, Total number of bacteria cells for Day 7 = 1850.3301941 + 2256.5002367 = 4106.8304308 cells

Approximately to nearest whole number, the total number of bacteria cells that would be present after 7 days = 4107 cells

3 0

1 year ago

Mr. Jackson invested a sum of money at 6% per year, and 3 times as much at 4%

bogdanovich [222]

Mr. Jackson invested $800 at 6% per year and $ 2400 at 4 % per year

<h3><u>Solution:</u></h3>

Mr. Jackson invested a sum of money at 6% per year, and 3 times as much at 4% per year.

Let the sum invested be ‘a’ and ‘3a’ at 6% per year and 4 % per year respectively

Also, his annual return totaled $144

We can form following equation on the basis of question:-

$\begin{array}{l}{\text { Then, } \frac{a \times 6 \times 1}{100}+\frac{3 a \times 4 \times 1}{100}=\$ 144} \\\\ {\frac{6 a}{100}+\frac{12 a}{100}=144} \\\\ {\frac{6 a+12 a}{100}=144} \\\\ {\frac{18 a}{100}=144} \\\\ {18 a=14400} \\\\ {a=14400 \div 18}\end{array}$

a = $800

The amount of money invested at 6% = a = 800

The amount of money invested at 4 % = 3a = 3(800) = 2400

So, the amount of money invested at 6% is $800 and the amount of money invested at 4% is $ 2400

4 0

1 year ago

1. One student was upset with her grade on Test 2 and thought the test was too difficult. In her anger, she said, "There is no w

Lemur [1.5K]

Answer:

Hypothesis rejected

Step-by-step explanation:

Lets use the t-test since the variance of the population is now known. We need to test the hypothesis that H_0: \mu \leq 79 \text{ vs } H_1: \mu > 79 . This is performed in R as follows:

t.test(tt$Test.2.Score,mu=79,alternative="greater")

One Sample t-test

data: tt$Test.2.Score

t = 2.9238, df = 69, p-value = 0.002337

alternative hypothesis: true mean is greater than 79

95 percent confidence interval:

81.26555 Inf

sample estimates:

mean of x

84.27143

Thus, we reject the null hypothesis and conclude that \mu > 79.

4 0

2 years ago

which represents the solution(s) of the system of equations, y=x^2-2x-15 and y=8x-40? determine the solution set algebraically

Harrizon [31]

y=x^2-2x-15 (1) y=8x-40 (2)

8x-40=x^2-2x-15
X^2+10x-25=0
(x-5)^2
×=5 y=0

7 0

2 years ago

An experienced carpenter can frame a house twice as fast as an apprentice. Working together, it takes the carpenters 2 days. How

AleksAgata [21]

Answer:

Step-by-step explanation:

Let the time taken by Carpenter working alone = $C$ days

Then time taken by apprentice alone = Twice as that of taken by Carpenter = 2 $C$ days

Time taken working together = 2 days

Work done in one day working together = $\frac{1}{2}$

Work done in one day by Carpenter working alone = $\frac{1}{C}$

Work done in one day by apprentice working alone = $\frac{1}{2C}$

Work done in one day by Carpenter working alone + Work done in one day by Carpenter working alone = $\frac{1}{C}$ + $\frac{1}{2C}$ = Work done in one day working together = $\frac{1}{2}$

$\dfrac{1}{C}+\dfrac{1}{2C}=\dfrac{1}{2}\\\Rightarrow \dfrac{2+1}{2C}=\dfrac{1}{2}\\\Rightarrow C = 3\ days$

Time taken by Carpenter alone to complete the work = 3 days

Time taken by Apprentice alone to complete the work = 3 $\times$ 2= <em>6 days</em>

4 0

2 years ago