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8_murik_8 [283]
3 years ago
9

A scale drawing of a kitchen is shown below. The scale is 1:20.

Mathematics
1 answer:
Viefleur [7K]3 years ago
7 0

Answer:

4×20=80×2=160

5×20=100×2=200

160+200=360

area=360

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I need an equation that has variables on both sides were x = 17
Andrei [34K]

3 (3 + x) = 5x - 25

9 + 3x = 5x - 25

Add 25 to each side

34 + 3x = 5x

Subtract 3x from each side

34 = 2x

Divide each side by 2

17 = x

5 0
3 years ago
If two cards are randomly drawn from a standard 52-card deck, what is the probability that both cards are hearts?
Anettt [7]

Answer:

26

Step-by-step explanation:

You have to divide the two numbers, which are 2 and 52

5 0
3 years ago
Help once more please
REY [17]

Answer:

84

Step-by-step explanation:

they are all equal sides so what you'd do is 28+28+28= 84

5 0
3 years ago
Read 2 more answers
Simplify the expression using trigonometric identities: sec (–θ) – cos θ.
ololo11 [35]

Answer:

sin θ . tan θ

Step-by-step explanation:

Note : -

sec ( - θ ) = sec θ

Formula / Identity : -

sec θ = 1 / cos θ

sec ( - θ ) - cos θ

= [ 1 / cos θ ] - cos θ

{ LCM = cos θ }

= [ 1 / cos θ ] - [ cos²θ / cos θ ]

= [ 1 - cos²θ ] / cos θ

{ 1 - cos²θ = sin²θ }

= sin²θ / cos θ

{ sin²θ = sin θ . sin θ }

= sin θ . sin θ / cos θ

{ sin θ / cos θ = tan θ }

= sin θ . tan θ

Hence, simplified.

4 0
2 years ago
The concentration of particles in a suspension is 50 per mL. A 5 mL volume of the suspension is withdrawn. a. What is the probab
kolezko [41]

Answer:

(a) 0.6579

(b) 0.2961

(c) 0.3108

(d) 240

Step-by-step explanation:

The random variable <em>X</em> can be defined as the number of particles in a suspension.

The concentration of particles in a suspension is 50 per ml.

Then in 5 mL volume of the suspension the number of particles will be,

5 × 50 = 250.

The random variable <em>X</em> thus follows a Poisson distribution with parameter, <em>λ</em> = 250.

The Poisson distribution with parameter λ, can be approximated by the Normal distribution, when λ is large say λ > 10.  

The mean of the approximated distribution of X is:

μ = λ  = 250

The standard deviation of the approximated distribution of X is:

σ = √λ  = √250 = 15.8114

Thus, X\sim N(250, 250)

(a)

Compute the probability that the number of particles withdrawn will be between 235 and 265 as follows:

P(235

                             =P(-0.95

Thus, the value of P (235 < <em>X</em> < 265) = 0.6579.

(b)

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.28

Thus, the value of P(48.

(c)

A 10 mL sample is withdrawn.

Compute the probability that the average number of particles per mL in the withdrawn sample is between 48 and 52 as follows:

P(48

                             =P(-0.40

Thus, the value of P(48.

(d)

Let the sample size be <em>n</em>.

P(48

                             0.95=P(-z

The value of <em>z</em> for this probability is,

<em>z</em> = 1.96

Compute the value of <em>n</em> as follows:

z=\frac{\bar X-\mu}{\sigma/\sqrt{n}}\\\\1.96=\frac{48-50}{15.8114/\sqrt{n}}\\\\n=[\frac{1.96\times 15.8114}{48-50}]^{2}\\\\n=240.1004\\\\n\approx 241

Thus, the sample selected must be of size 240.

5 0
3 years ago
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