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Lyrx [107]
3 years ago
8

1 + 4 x 3 + 1 x 5 = 66 this statement is false. How to use parentheses to fix this?

Mathematics
1 answer:
Nezavi [6.7K]3 years ago
7 0
I got no clue, I tried many ways and didn’t get it
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What is the volume of 6f and 7ft and 9fr
NeX [460]
The volume is 378 ft³. Hope this helps! Good luck.
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3 years ago
Coach Williams bought trophies for his players at the end of the season. Each trophy is packed in a cube-shaped box that has a v
Diano4ka-milaya [45]

Answer:

The volume of the cart will be 24 ft^3

Step-by-step explanation:

The first thing we need to calculate for this question is the number of boxes that fill the entire cart.

A layer of boxes consists of 8 boxes.

The cart holds a maximum of 3 layers of boxes.

So, the total number of boxes held by the cart are:

Total boxes = number of layers * boxes per layer

Total boxes = 3 * 8

Total boxes = 24

Since each box has a volume of one cubic foot, the total volume of the cart will be:

Volume of cart = number of boxes * volume of each box

Volume of cart = 24 * 1

Volume of cart = 24 ft^3

5 0
3 years ago
A zoologist is studying four very closely related feline species. She wishes to compare their gestation periods. An observationa
diamong [38]

Answer:

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different, NOT to conclude that the all the means are different.

Step-by-step explanation:

Previous concepts

Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".  

The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"  

Solution to the problem

The hypothesis for this case are:

Null hypothesis: \mu_{A}=\mu_{B}=\mu_{C}= \mu_D

Alternative hypothesis: Not all the means are equal \mu_{i}\neq \mu_{j}, i,j=A,B,C,D

In order to find the mean square between treatments (MSTR), we need to find first the sum of squares and the degrees of freedom.

If we assume that we have p=4 groups and on each group from j=1,\dots,p we have n_j individuals on each group we can define the following formulas of variation:  

SS_{total}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x)^2  

SS_{between}=SS_{model}=\sum_{j=1}^p n_j (\bar x_{j}-\bar x)^2  

SS_{within}=SS_{error}=\sum_{j=1}^p \sum_{i=1}^{n_j} (x_{ij}-\bar x_j)^2  

And we have this property  

SST=SS_{between}+SS_{within}  

And in order to test this hypothesis we need to ue an F statistic and for this case the p value calculated is

p_v= 0.01

Since the significance level is 0.05 we see that pv so we have enough evidence to reject the null hypothesis. And the best conclusion for this case would be:

b. at least some, but not all, of the gestation periods across all four species are the same

Because is only to identify if AT LEAST one mean is different NOT to conclude that the all the means are different.

8 0
4 years ago
Store
4vir4ik [10]
The answer to this is D , 2.02$ i think
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3 years ago
Cost of admission to a town fair was $8 for adults and $5 for children. If 206 people
tatiyna
I think 176 adult tickets were sold , sry if it’s wrong i’m kinda bad in math
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3 years ago
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