Please help it’s due by tonight! The other two were posted if you are confused so you can see the full problem

Mondays at pizza pan you can cheese to toppings at no extra cost for your pizza to tacos are mushroom sausage peppers and extra

viva [34]

Answer:

We can order 35 different types of pizza based on the toppings.

Step-by-step explanation:

There are 3 different types of pizza crust which a person can choose for his pizza. On Monday's company has introduced new campaign in which a person can choose extra toppings of his choice at no additional cost. The topping are of 7 different types. The total number of different pizza which a person can have is 7C3. Solving through computation technique we get 35.

7 0

2 years ago

The diameter of a circle is 1 meter.the circumference is how many centimeters

andre [41]

Step-by-step explanation:

C= pi×d

d=.5 meters

so plug in the numbers

3.14 × .5= 1.75

circumference= 1.75 meters

3 0

2 years ago

A random sample of 64 observations produced a mean value of 84 and standard deviation of 5.5. The 90% confidence interval for th

ASHA 777 [7]

Answer: The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

Step-by-step explanation:

When population standard deviation is not given ,The confidence interval population proportion is given by ( $\mu$ ):-

$\overline{x}\pm t^*\dfrac{s}{\sqrt{n}}$

, where n= Sample size.

s= Sample standard deviation

$\overline{x}$ = sample mean

t* = Critical t-value (Two-tailed)

As per given , we have

$\overline{x}=84$

n= 64

Degree of freedom : df = n-1=63

s= 5.5

Significance level : $\alpha=1-0.90=0.1$

Two-tailed T-value for df = 63 and $\alpha=1-0.90=0.1$ would be

$t_{\alpha/2,df}=t_{0.05,63}=1.669$ (By t-distribution table)

i.e. t*= 1.669

The 90% confidence interval for the population mean μ would be

$84\pm (1.669)\dfrac{5.5}{\sqrt{64}}$

$=84\pm (1.669)\dfrac{5.5}{8}$

$\approx84\pm 1.15$

$=(84-1.15,\ 84+1.15)=(82.85,\ 85.15)$

∴ The 90% confidence interval for the population mean μ is between 82.85 and 85.15,

6 0

3 years ago

Help guys please. I

tatuchka [14]

Step-by-step explanation:

you need to use the ruler to measure the lengths of it, like place the ruler from A to B and look at the measurement of it

7 0

1 year ago

Use Lagrange multipliers to find the maximum and minimum values of (i) f(x,y)-81x^2+y^2 subject to the constraint 4x^2+y^2=9. (i

sp2606 [1]

i. The Lagrangian is

$L(x,y,\lambda)=81x^2+y^2+\lambda(4x^2+y^2-9)$

with critical points whenever

$L_x=162x+8\lambda x=0\implies2x(81+4\lambda)=0\implies x=0\text{ or }\lambda=-\dfrac{81}4$

$L_y=2y+2\lambda y=0\implies2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_\lambda=4x^2+y^2-9=0$

If $x=0$ , then $L_\lambda=0\implies y=\pm3$ .
If $y=0$ , then $L_\lambda=0\implies x=\pm\dfrac32$ .
Either value of $\lambda$ found above requires that either $x=0$ or $y=0$ , so we get the same critical points as in the previous two cases.

We have $f(0,-3)=9$ , $f(0,3)=9$ , $f\left(-\dfrac32,0\right)=\dfrac{729}4=182.25$ , and $f\left(\dfrac32,0\right)=\dfrac{729}4$ , so $f$ has a minimum value of 9 and a maximum value of 182.25.

ii. The Lagrangian is

$L(x,y,z,\lambda)=y^2-10z+\lambda(x^2+y^2+z^2-36)$

with critical points whenever

$L_x=2\lambda x=0\implies x=0$ (because we assume $\lambda\neq0$ )

$L_y=2y+2\lambda y=0\implies 2y(1+\lambda)=0\implies y=0\text{ or }\lambda=-1$

$L_z=-10+2\lambda z=0\implies z=\dfrac5\lambda$

$L_\lambda=x^2+y^2+z^2-36=0$

If $x=y=0$ , then $L_\lambda=0\implies z=\pm6$ .
If $\lambda=-1$ , then $z=-5$ , and with $x=0$ we have $L_\lambda=0\implies y=\pm\sqrt{11}$ .

We have $f(0,0,-6)=60$ , $f(0,0,6)=-60$ , $f(0,-\sqrt{11},-5)=61$ , and $f(0,\sqrt{11},-5)=61$ . So $f$ has a maximum value of 61 and a minimum value of -60.

5 0

2 years ago