Answer:
S varies partly directly as M and Q.
S=C.
S=KMQ+C.
For the first one...
speed=80,m=220,Q=30.
80=K20×30+C.
80=600K+C......(I).equation one.
For the second one....
speed=60,m=300,Q=40.
60=K300×40+C.
60=12000K+C.....(ii). equation two.
Minus eqtn(I) from eqtn(ii).
80=600K+C.
- 60=12000K+C.
K=0.01754~0.018.
Substitute K=0.018 into eqtn(I).
80=600K+C
80=600×0.018+C.
80=10.8+C.
C=80-10.8=69.2.
The relation is S=0.018MQ+69.2
when speed is 100 and mass is 250 find the volume.
100=0.018×250×Q+69.2.
100=4.5Q+69.2.
4.5Q=100-69.2
4.5Q=30.8.
Q=30.8/4.5.
Q=6.8~7litres.
Answer:
The correct option is;
c. Because the p-value of 0.1609 is greater than the significance level of 0.05, we fail to reject the null hypothesis. We conclude the data provide convincing evidence that the mean amount of juice in all the bottles filled that day does not differ from the target value of 275 milliliters.
Step-by-step explanation:
Here we have the values
μ = 275 mL
275.4
276.8
273.9
275
275.8
275.9
276.1
Sum = 1928.9
Mean (Average), = 275.5571429
Standard deviation, s = 0.921696159
We put the null hypothesis as H₀: μ₁ = μ₂
Therefore, the alternative becomes Hₐ: μ₁ ≠ μ₂
The t-test formula is as follows;
Plugging in the values, we have,
Test statistic = 1.599292
at 7 - 1 degrees of freedom and α = 0.05 = ±2.446912
Our p-value from the the test statistic = 0.1608723≈ 0.1609
Therefore since the p-value = 0.1609 > α = 0.05, we fail to reject our null hypothesis, hence the evidence suggests that the mean does not differ from 275 mL.
Answer:
ma ma ھگدفسخ टगाई samje ke hamreبھا मे
hoho gana ga
Answer:
36
Step-by-step explanation:
You simply multiply the number of options per each course of the meal and get 3*4*3=12*3=36
(a^3b)^2 • 4ab^3
a^6b • 4ab^3
4a^6b+1b^3