Answer:
18
Step-by-step explanation:
as there are 12 red spheres there are 12*4 red cubes (48) add them ,60.Then 60/2 to find blue shapes ,30, then add the 2 and 3 to make 5 after that do 30/5 = 6 then multiply by 3
Answer:
124.67cm³
Step-by-step explanation:
==>Given:
Dimensions of current can:
Height (h) = 12cm
Diameter = 6cm (radius = 3cm)
Volume of current can (V1) = 339.12 cm³
Dimensions of future cans to be increased multiple of 1.11:
height = 12cm × 1.11 = 13.32 cm
radius = 3cm × 1.11 = 3.33 cm
Volume of future can (V2) = πr²h = 3.14*3.33²*13.32
= 3.14*11.0889*13.32 = 463.791025
V2 ≈ 463.79 cm³
==>Find how much more would new cans hold = Volume of new can - volume of current can
= 463.79 cm³ - 339.12 cm³
= 124.67cm³
Answer:
B) 28.53 unit²
Step-by-step explanation:
The diagonal AD divides the quadrilateral in two triangles:
- Triangle ABD
- Triangle ACD
Area of Quadrilateral will be equal to the sum of Areas of both triangles.
i.e.
Area of ABCD = Area of ABD + Area of ACD
Area of Triangle ABD:
Area of a triangle is given as:

Base = AB = 2.89
Height = AD = 8.6
Using these values, we get:

Thus, Area of Triangle ABD is 12.43 square units
Area of Triangle ACD:
Base = AC = 4.3
Height = CD = 7.58
Using the values in formula of area, we get:

Thus, Area of Triangle ACD is 16.30 square units
Area of Quadrilateral ABCD:
The Area of the quadrilateral will be = 12.43 + 16.30 = 28.73 units²
None of the option gives the exact answer, however, option B gives the closest most answer. So I'll go with option B) 28.53 unit²
Answer:
Check the explanation
Step-by-step explanation:
One way ANOVA
The null and alternative hypothesis for this one way ANOVA is given as below:
Null hypothesis: H0: There is no significant difference in the averages of the scores for the quizzes, exams and final only.
Alternative hypothesis: There is a significance difference in the averages of the scores for the quizzes, exams and final only.
The ANOVA table with calculations can be seen in the attached images below:
In the attached image below, we get the p-value for this one way ANOVA test as 0.0221. We do not reject the null hypothesis if the p-value is greater than the given level of significance and we reject the null hypothesis if the p-value is less than the given level of significance or alpha value.
In the attached image below, we are given that the p-value = 0.0221 and level of significance or alpha value = 0.05, that is p-value is less than the given level of significance. So, we reject the null hypothesis that there is no significant difference in the averages of the scores for the quizzes, exams and final only. This means we conclude that there is a significance difference in the averages of the scores for the quizzes, exams and final only.