33.32 divided by 9.8

A fish market bought two swordfish at a rate of $13 per pound the cost of the larger fish was 3 times as great as the cost of th

Orlov [11]

3952/13 is 304 pounds at the cost of 13 dollars each fish is 152 pounds

6 0

2 years ago

Four men are to divide K500 equally among them. When the money was given, 20% was taken away.

Agata [3.3K]

Answer: 20% of 500= 100

So 500-100 = 400

4x100= 400

Step-by-step explanation:

4 0

2 years ago

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

2 years ago

The sum of 5x2y and (2xy2 + x2y) is

allochka39001 [22]

5x²y + 2xy² + x²y

Combining like terms would be
6x²y + 2xy²

The two terms are now unique and cannot be combined any further.

6 0

2 years ago

A sociologist wants to determine if the life expectancy of people in africa is less than the life expectancy of people in asia.

Colt1911 [192]

Answer:

The sample statistics are attached.

First, we need to determine the hypothesis.

The null hypothesis should be: $H_{o} : u_{africa} \geq u_{asia}$ .

The alternative hypothesis should be: $H_{a}: u_{africa}$

The next step is about calculating the t statistics based on the samples. This t test will give the probability value or p-value, which determines if there's enough to reject the null hypothesis. The formula we need to use to find the t-value is: $t=\frac{x_{1}-x_{2} }{\sqrt{\frac{s_{1} ^{2} }{n_{1} } +\frac{s_{2} ^{2} }{n_{2} } } }$

Replacing all the values into the formula, we have:

$t=\frac{52.4-58.1}{\sqrt{\frac{(7.4)^{2} }{45}+\frac{(8.8)^{2} }{35} } } \\t=\frac{-5.7}{\sqrt{1.2+2.2} } =\frac{-5.7}{1.8} =-3.2$

Therefore, the t-value is -3.2.

Now, we using a level of significance of 0.01, and a degree of freedom (df) of 34, we use the t-table to find the p-value for this results. (df = N -1; in this case, we take the smaller sample, which is 35, giving us 34 of df).

Therefore, according to the table attached, the p-value is less than 0.01, which is less than our level of significance. This result means that the null hypothesis is rejected. In other words, there's enough evidence to say that the life expectancy of people in Africa is less than the life of expectancy of people in Asia.

3 0

2 years ago