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o-na [289]
1 year ago
7

Evaluate the expression for x = 2 and y = 4. Enter your answer in the box. 16x0 + 2x2 • y –1

Mathematics
1 answer:
swat321 year ago
6 0

Given that,

The given expression is : 16x^0+2x^2{\cdot}y^{-1}

To find,

The value of the above expression when x = 2 and y = 4

Solution,

We have,

16x^0+2x^2{\cdot}y^{-1}

Put x = 2 and y = 4 in the above expression.

16x^0+2x^2{\cdot}y^{-1}\\\\=16\times (2)^0+2(2)^2\times (4)^{-1}\\\\=16\times 1+2(2)^2\times\dfrac{1}{4}\\\\=16+2\\\\=18

So, the value of the above expression is 18.

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Jan accidentally ran 7 minutes longer than she was supposed to. Write an expression for the total amount of time Jan ran if she
vovangra [49]

Answer:

m+7

Step-by-step explanation:

If she ran 7 minutes longer and m represents the amount of time she ran, then it would just be addition.

Let's say m=60 minutes. She would've ran for 67 minutes, which is also m+7.

6 0
2 years ago
Find three different surfaces that contain the curve r(t) = t^2 i + lnt j + (1/t)k
Viefleur [7K]

solution:

Consider the curve: r(t) = t²i +(int)j + 1/t k

X= t² , y = int ,z = 1/t

Using, x = t², z = 1/t

                    X = (1/z)²

Xz²= 1

Using y = int, z= 1/t

Y = in│1/z│

Using x = t², y = int

Y = int

= in(√x)

Hence , the required surface are,

Xz² = 1

Y = in│1/z│

Y= in(√x)


7 0
2 years ago
If m angle BGF=152 degrees, what is m angle AGF
Allushta [10]

The Answer is: 64 degrees

4 0
2 years ago
Read 2 more answers
A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
SVEN [57.7K]

Answer:

Step-by-step explanation:

Hello!

X: number of absences per tutorial per student over the past 5 years(percentage)

X≈N(μ;σ²)

You have to construct a 90% to estimate the population mean of the percentage of absences per tutorial of the students over the past 5 years.

The formula for the CI is:

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

⇒ The population standard deviation is unknown and since the distribution is approximate, I'll use the estimation of the standard deviation in place of the population parameter.

Number of Absences 13.9 16.4 12.3 13.2 8.4 4.4 10.3 8.8 4.8 10.9 15.9 9.7 4.5 11.5 5.7 10.8 9.7 8.2 10.3 12.2 10.6 16.2 15.2 1.7 11.7 11.9 10.0 12.4

X[bar]= 10.41

S= 3.71

Z_{1-\alpha /2}= Z_{0.95}= 1.645

[10.41±1.645*(\frac{3.71}{\sqrt{28} } )]

[9.26; 11.56]

Using a confidence level of 90% you'd expect that the interval [9.26; 11.56]% contains the value of the population mean of the percentage of absences per tutorial of the students over the past 5 years.

I hope this helps!

7 0
2 years ago
Each airline passenger and his or her luggage must be checked to determine whether he or she is carrying weapons on to the airpl
Dafna1 [17]

Answer:

a

P_k  = 0.83

b

 N_{\mu} \approx  4 \ passengers

c

T_{\lambda} =  0.5 \ minutes

Step-by-step explanation:

From the question we are told that

The average number of passengers that arrive per minute is \lambda = 10

The average number of check that can be carried out in one minute is \mu= 12

Generally the probability that a passenger will have to wait before being checked for weapons is mathematically represented as

        P_k  = \frac{\lambda }{ \mu }

=>    P_k  = \frac{10 }{ 12}

=>    P_k  = 0.83

Generally the number of passengers are waiting in line to enter the checkpoint is mathematically represented as

     N_{\mu} =  \frac{\lambda^2}{\mu (\mu -\lambda) }

=>  N_{\mu} =  \frac{10^2}{12 (12 -10) }

=>  N_{\mu} \approx  4 \ passengers

Generally the average time a passenger spend at the checkpoint is mathematically represented as

      T_{\lambda} = \frac{ \frac{\lambda}{(\mu - \lambda)} }{ \lambda}

=>   T_{\lambda} = \frac{ \frac{ 10}{(12 - 10)} }{10}

=>   T_{\lambda} =  0.5 \ minutes

5 0
1 year ago
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