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3 years ago

I really need help please

Mathematics

2 answers:

dem82 [27]3 years ago

7 0

0.3 for the first one
0.4 for the second

Ahat [919]3 years ago

4 0

Answer:

the first one and the fith one

Step-by-step explanation:

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A manufacturer is running an experiment using the standard (control) infant formula, and two new formulas, A and B. The goal is

bulgar [2K]

Answer:

H0: the mean IGA score is the same for all three formulas. 

Ha: the mean IGA score is not the same for all three formulas.

The Group of answer choices are as below:

H0: the mean IGA score is the same for all three formulas. 

Ha: the mean IGA score is not the same for all three formulas.

H0: the mean IGA score is the same for all three formulas. 

Ha: the mean IGA score is higher for at least one of the two treatment groups than the control group

H0: the mean IGA score is the same for all three formulas. 

Ha: the mean IGA score is higher for both treatment groups A and B than the control group.

D. No answer text provided.

Explanation:

Given that a manufacturer is running an experiment using the standard (control) infant formula, and two new formulas, A and B to boost the immune system in infants.

The variable that is measured is Total IGA (in mg per dl) score.

We have to state the hypotheses for one-way ANOVA F test to test if the mean IGA score is the same for all three formulas.

In null hypothesis H0, the mean of all groups are equal and in alternative hypothesis, Ha, the mean of the groups are not the same.

Therefore, the hypothesis tested by the one-way ANOVA F test are:

H0: the mean IGA score is the same for all three formulas.

Ha: the mean IGA score is not the same for all three formulas.

5 0

3 years ago

Help with the question above! Pleaseeeeee!

SSSSS [86.1K]

Question 5:
the circumference is given by:
C = 2 * pi * r
Where,
r: radio of the ball
Substituting values we have:
22 = 2 * pi * r
Clearing r we have:
r = 11 / pi
The surface area is given by:
A = 4 * pi * r ^ 2
Substituting values we have:
A = 4 * 3.14 * (11 / 3.14) ^ 2
A = 154 in ^ 2
Answer:
The surface area of the balloon is:
A = 154 in ^ 2

Question 8:
For this case we have that the scale factor is given by:
V1 = (k ^ 3) * V2
Substituting values we have:
729 = (k ^ 3) * 2744
Clearing k:
k = (729/2744) ^ (1/3)
k = 9/14
Answer:
the scale factor of a cube with volume 729 m ^ 3 to a cube with volume 2,744 m ^ 3 is:
9:14

Question 2:
The volume of the cylinder is given by:
V = pi * r ^ 2 * h
Where,
r : radio
h: height
Substituting values:
V = pi * (2.8) ^ 2 * (13)
V = 101.92 * pi
Answer:
The volume of the cylinder is:
V = 101.92 * pi
option 3

4 0

3 years ago

Read 2 more answers

A jar contains 17 yellow and 12 green marbles. You randomly draw one marble from the jar then

dexar [7]

Answer:

204/841

Step-by-step explanation:

17/29 times 12/29

3 0

2 years ago

If the volume of the solid shown in the figure is 438 cubic inches, find the value of x.

svetoff [14.1K]

Answer:

what

Step-by-step explanation:

what figure

7 0

1 year ago

Find [5(cos 330 degrees + I sin 330 degrees)]^3

earnstyle [38]

Given a complex number in the form:
$z= \rho [\cos \theta + i \sin \theta]$
The nth-power of this number, $z^n$ , can be calculated as follows:

- the modulus of $z^n$ is equal to the nth-power of the modulus of z, while the angle of $z^n$ is equal to n multiplied the angle of z, so:
$z^n = \rho^n [\cos n\theta + i \sin n\theta ]$
In our case, n=3, so $z^3$ is equal to
$z^3 = \rho^3 [\cos 3 \theta + i \sin 3 \theta ] = (5^3) [\cos (3 \cdot 330^{\circ}) + i \sin (3 \cdot 330^{\circ}) ]$ (1)
And since
$3 \cdot 330^{\circ} = 990^{\circ} = 2\pi +270^{\circ}$
and both sine and cosine are periodic in $2 \pi$ , (1) becomes
$z^3 = 125 [\cos 270^{\circ} + i \sin 270^{\circ} ]$

6 0

3 years ago