Answer:
It is likely that the birth weight of a random baby boy will be between 3.2 and 3.4 kg because the probability of this event is large enough.
Step-by-step explanation:
Population mean=μ=3.3.
S.E=0.1.
n=36.
If the probability of the birth weight of a random baby boy will be between 3.2 and 3.4 kg is larger than the it will be likely. The probability can be calculated by normal distribution because sample size is large enough.
Z-score for 3.2 kg=3.2-3.3/0.1=-1
Z-score for 3.4 kg=3.4-3.3/0.1=1
P(-1<Z<1)=P(-1<Z<0)+P(0<Z<1)
P(-1<Z<1)=0.3413+0.3413
P(-1<Z<1)=0.6826
The probability of the birth weight of a random baby boy will be between 3.2 and 3.4 kg is 68.26%. So. it is likely that the birth weight of a random baby boy will be between 3.2 and 3.4 kg as the probability is large enough.
Answer:
22 weeks
Step-by-step explanation:
it is 22 weeks because if you divide 39 by 2 then it is 19.5 and if you divide 429 by 19.5 you will get 22 hope this helps.
Answer:
5 small containers will be required and used.
This water is used to fill four and half small containers exactly.
4.5 small containers
Step-by-step explanation:
The rectangular container has dimensions 12*12*45 cm and the small container has dimensions 6*5*8 cm
The rectangular container with a square base is 2/3 full of water.The height is used up to 45*2/3= 7.5 cm
The volume of the water in the rectangular container with a square base is
12*12*7.5= 1080 cm³
The volume of the water in the smaller container is
6*5*8= 240 cm³
Number of smaller containers required = Volume of large container / Volume of smaller container
Number of smaller containers required =1080 cm³/240 cm³= 4.5
This water is used to fill four and half small containers exactly.
9514 1404 393
Answer:
105.0°, 255.0°
Step-by-step explanation:
Many calculators do not have a secant function, so the cosine relation must be used.
sec(θ) = -3.8637
1/cos(θ) = -3.8637
cos(θ) = -1/3.8637
θ = arccos(-1/3.8637) ≈ 105.000013°
The secant and cosine functions are symmetrical about the line θ = 180°, so the other solution in the desired range is ...
θ = 360° -105.0° = 255.0°
The angles of interest are θ = 105.0° and θ = 255.0°.
Answer:
He has 0 yards
Step-by-step explanation:
Initially he lost 10 yards
This means that he is 10 yards away from the starting point which we shall call origin
Afterwards, he gained 10 yards
what this mean is that he is back at the original point before the loss
Since he gained what he lost, then he is back at the origin and then we can say that no yards was lost or gain so we have him gaining 0 yards only