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2 years ago

21.05 divided by 0.2

Mathematics

2 answers:

Valentin [98]2 years ago

8 0

Answer:

Step-by-step explanation:

Aleks [24]2 years ago

6 0

Answer:

$\frac{21.05}{0.2}=105.25$

Step-by-step explanation:

you have to do long division, but there is a number of steps, too many to put in here, but just check on calculator if you want to make sure

You might be interested in

Traffic speed: The mean speed for a sample of cars at a certain intersection was kilometers per hour with a standard deviation o

aliina [53]

Answer:

Step-by-step explanation:

Hello!

X₁: speed of a motorcycle at a certain intersection.

n₁= 135

X[bar]₁= 33.99 km/h

S₁= 4.02 km/h

X₂: speed of a car at a certain intersection.

n₂= 42 cars

X[bar]₂= 26.56 km/h

S₂= 2.45 km/h

Assuming

X₁~N(μ₁; σ₁²)

X₂~N(μ₂; σ₂²)

and σ₁² = σ₂²

<em>A 90% confidence interval for the difference between the mean speeds, in kilometers per hour, of motorcycles and cars at this intersection is ________.</em>

The parameter of interest is μ₁-μ₂

(X[bar]₁-X[bar]₂)± $t_{n_1+n_2-2}$ * $Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} }$

$t_{n_1+n_2-2;1-\alpha /2}= t_{175; 0.95}= 1.654$

$Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{134*16.1604+41*6.0025}{135+42-2} } = 3.71$

[(33.99-26.56) ± 1.654 *( $3.71*\sqrt{\frac{1}{135} +\frac{1}{42} }$ )]

[6.345; 8.514]= [6.35; 8.51]km/h

<em>Construct the 98% confidence interval for the difference μ₁-μ₂ when X[bar]₁= 475.12, S₁= 43.48, X[bar]₂= 321.34, S₂= 21.60, n₁= 12, n₂= 15</em>

$t_{n_1+n_2-2;1-\alpha /2}= t_{25; 0.99}= 2.485$

$Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{11*(43.48)^2+14*(21.60)^2}{12+15-2} } = 33.06$

[(475.12-321.34) ± 2.485 *( $33.06*\sqrt{\frac{1}{12} +\frac{1}{15} }$ )]

[121.96; 185.60]

I hope this helps!

3 0

2 years ago

Six teachers and 12 students volunteer for a committee to discuss extra-curricular activities. How many committees of 5 people c

Alex17521 [72]

Answer:

a) $X=3300$

b) $Y=7770$

Step-by-step explanation:

From the question we are told that:

Number Teachers $T=6$

Number Student $S=12$

Number in committee $n=5$

a) Generally the equation for exactly 3 students on the committee is mathematically given by

$X=^{S}C_3*^{T}C_3$

$X=^{12}C_3*^{6}C_3$

$X=3300$

b) Generally the equation for at least one teacher and at least one student on the committee is mathematically given by

Total Ways-(no of ways of selection no teacher or student)

Where total Ways

$T=^{(6+12)}C_5$

$T=8568$

Therefore

$Y=8568-^{6}C_0*^{12}C_5+^{12}C_0*^{6}C_5$

$Y=8568-798$

$Y=7770$

8 0

2 years ago

A contractor has submitted bids on three state jobs: an office building, a theater, and a parking garage. State rules do not all

german

Answer:

The following are the answer to this question:

Step-by-step explanation:

In the given question the numeric value is missing which is defined in the attached file please fine it.

Calculating the probability of the distribution for x:

$\to f(x) = 0.19\ for \ x=14\\\\\to f(x) = 0.29 \ for\ x=7\\\\\to f(x) = 0.38\ for \ x=1\\\\\to f(x)=0.14 \ for \ x=0\\$

The formula for calculating the mean value:

$\bold{ E(X)= x \times f(x)}$

$=14 \times 0.19+7 \times 0.29+1 \times 0.38+0\times 0.14\\\\=2.66 + 2.03+0.38+ 0\\\\=5.07$

$\bold{E(X^2) = x^2 \times f(x)}$

$=14^2 \times 0.19+7^2 \times 0.29+1^2 \times 0.38+0^2 \times 0.14 \\\\=196 \times 0.19+ 49 \times 0.29+1 \times 0.38+0 \times 0.14\\\\= 37.24+ 14.21+ 0.38+0 \\\\=51.83$

use formula for calculating the Variance:

$\to \bold{\text{Variance}= E(X^2) -[E(X)]^2}$

$= 51.83 - (5.07)^2\\\\= 51.83 - 25.70\\\\=26.13$

calculating the value of standard deivation:

Standard Deivation (SD) = $\sqrt{Variance}$

$= \sqrt{26.13} \\\\=5.111$

6 0

2 years ago

Rammy has $9.60 to spend on some peaches and a gallon of milk. Peaches

sergeinik [125]

Answer:

$\large \boxed{\text{5.00 lb}}$

Step-by-step explanation:

$\begin{array}{rcl}1.20x + 3.60 & \leq & 9.60\\1.20x & \leq & 6.00\\x &\leq & \mathbf{5.00}\\\end{array}\\\text{Rammy can buy $\large \boxed{\textbf{5.00 lb}}$ of peaches.}$