Find the surface area of the rectangular prism.

Data was collected for 300 fish from the North Atlantic. The length of the fish (in mm) is summarized in the GFDT below.

agasfer [191]

Answer:

Class Boundary = 1 between the sixth and seventh classes.

Step-by-step explanation:

Lengths (mm) Frequency

1. 140 - 143 1

2. 144 - 147 16

3. 148 - 151 71

4. 152 - 155 108

5. 156 - 159 83

6. 160 - 163 18

7. 164 - 167 3

The class boundary between two classes is the numerical value between the starting value of the higher class, which is 164 for the 7th class in this case, and the ending value of the class of the lower class, which is 163 for the 6th class in this case.

Therefore the class boundary between the sixth and seventh classes

= 164 - 163 = 1

Therefore Class Boundary = 1.

It can be seen that class boundary for the frequency distribution is 1.

If we take the difference between the lower limits of one class and the lower limit of the next class then we will get the class width value.

Therefore, Class width,

w = lower limit of second class - lower limit of first class

= 144 - 140

= 4

7 0

2 years ago

Gibbs Baby Food Company wishes to compare the weight gain of infants using its brand versus its competitor’s. A sample of 40 bab

Leviafan [203]

Answer:

$z=\frac{(7.6-8.1)-0}{\sqrt{\frac{2.3^2}{40}+\frac{2.9^2}{55}}}}=-0.936$

The p value can be founded with this formula:

$p_v =P(z$

Since the p value is higher than the significance level provided of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean for the Gibbs brand is significantly lower than the true mean for the competitor

Step-by-step explanation:

Information given

$\bar X_{1}=7.6$ represent the mean for Gibbs products

$\bar X_{2}=8.1$ represent the mean for the competitor

$\sigma_{1}=2.3$ represent the population standard deviation for Gibbs

$\sigma_{2}=2.9$ represent the sample standard deviation for the competitor

$n_{1}=40$ sample size for the group Gibbs

$n_{2}=55$ sample size for the group competitor

$\alpha=0.05$ Significance level provided

z would represent the statistic

Hypothesis to verify

We want to check if babies using the Gibbs brand gained less weight, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}< 0$

The statistic would be given by:

$z=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

Replacing the info given we got:

$z=\frac{(7.6-8.1)-0}{\sqrt{\frac{2.3^2}{40}+\frac{2.9^2}{55}}}}=-0.936$

The p value can be founded with this formula:

$p_v =P(z$

Since the p value is higher than the significance level provided of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can't conclude that the true mean for the Gibbs brand is significantly lower than the true mean for the competitor

5 0

2 years ago

Evaluate 3ab for a=2 and b=3

vampirchik [111]

Answer:

18

Step-by-step explanation:

For this problem, we will simply plug in the values of a and b into the respective variables in the expression 3ab to "evaluate" the expression.

a = 2; b = 3

3ab

Note, that when variables like a and b are smashed with a constant, the use of multiplication is in play.

3ab = 3 * a * b

With this in mind, let's plug in the values of a and b into the expression.

3ab

= 3 (2) (3)

= 3 (6)

= 18

Hence, 3ab evaluated when a=2 and b=3 is 18.

Cheers.

4 0

2 years ago

Read 2 more answers

Indicate whether the function represents an exponential growth or decay.

svetoff [14.1K]

Answer:

1) decay

2) growth

3) growth

Step-by-step explanation:

A generic exponential function can be written as:

f(x) = A*(r)^x

Where:

A is the initial amount of something.

r is the rate of growth.

x is the variable, usually, represents time.

if r > 1, we have an exponential growth.

if r < 1, we have an exponential decay.

1) f(x) = (3/4)^x

in this case we have:

A = 1

r = (3/4) = 0.75

Clearly, r < 1.

Then this is an exponential decay.

2) f(x) = (1/6)*4^x

In this case we have:

A = (1/6)

r = 4

Here we have r > 1.

Then this is an exponential growth.

3) f(x) = (1/4)*(5/2)^x

in this case we have:

A = 1/4

r = 5/2 = 2.5

here we have r > 1, then this is an exponential growth.

8 0

3 years ago

Plz plz help me so i can get a good grade

Levart [38]

Area of the figure = 806.5 in²

Solution:

Length of the rectangle = 16 in

Breadth of the rectangle = 9 in

Area of the rectangle = length × breadth

= 16 × 9

Area of the rectangle = 144 in²

Base of the triangle = 31 in

Height of the triangle = 20 in

Area of the triangle = $\frac{1}{2}bh$

$$=\frac{1}{2}\times 31\times 20$

Area of the triangle = 310 in²

Parallel sides of the trapezium = 16 in and 31 in

Height of the trapezium = 35 – 20 = 15 in

Area of the trapezium = $\frac{1}{2}\times \text{Sum of the parallel sides}\times \text{Height}$

$$=\frac{1}{2}\times {(16+31)}\times{15}$

$$=\frac{1}{2}\times {47}\times{15}$

Area of the trapezium = 352.5 in²

Area of the figure = Area of rectangle + Area of triangle + Area of trapezium

= 144 in² + 310 in² + 352.5 in²

Area of the figure = 806.5 in²

8 0

3 years ago