What do I use to make a meat loaf?

Answers

Answer 1

Instructions

1. Preheat oven to 350 degrees Fahrenheit.
2. In a large bowl, add the beef, bread crumbs, onion, milk, egg, 2 tablespoons ketchup, worcestershire sauce, parsley, salt, garlic powder, and pepper. Use your hands to mush and mix these ingredients together until well combined.*
3. Add the meat mixture to a loaf pan. Pat the meat down into an even layer.
4. In a small bowl, add 1/4 cup ketchup, the brown sugar, and vinegar. Stir to combine. Pour the sauce on top of the meatloaf and spread it into an even layer.
5. Bake uncovered for 55 minutes.

Related Questions

What is the probability of scoring a total of 6

Answers

what are the choices

can you explain some more please

Translate subtract 10 from r, then subtract 4 from the result

Answers

Answer:

reste 10 de r, luego reste 4

Step-by-step explanation:Lets say the result is x x is a letter driving from spain so x=Spanish so translating that sentence to spanish is easy

Answer:

-r+6

Step-by-step explanation:

If the volleyball team washes 10 more cars, then they will have met their overall goal of
washing 60 cars to raise money for their season. Let c represent the number of cars that
the volleyball team has washed so far. Which of the following equations can be used to
solve for c?

Answers

Answer:

Step-by-step explanation:

60-10=50

they washed 50 cars.

Picture is it............

Answers

In this triangle the three interior angles are = ∠82° , ∠54° and ∠x .

Then ;

∠82° + ∠54° + ∠x = 180° ( under angle sum property which says that the sum of the three interior angles of a triangle is equal to 180° )

= 136 ° + x = 180

= x = 180 - 136

= x = 44°

∠x and ∠y will sum up to 180° as they are a linear pair .

= 44° + y = 180

= y = 180 - 44

= y = 136°

Since ∠y and ∠z are vertically opposite angles their angle measure will be equal . Which means ;

= ∠y = ∠z

= 136° = ∠z

= ∠z = 136°

Therefore :-

The value of x = 44°

Whats the function to this math prob

Answers

Answer:

some reason i cant see the image is this just me

Step-by-step explanation:

Y is the vertical distance. The top of the curve is at y = 4 and the bottom of the curve is at y = -5, so the function would be between y -5 and y 5 which is written as :

D. -5 <= y <= 5

How do you solve this? Without going into anything to complicated as this should be Year 10 maths

Answers

Given : ABCD is a square with each side 5cm .

To Find : The area of the shaded region .

Solution : On observing the figure we can see two quadrants , quadrant ADC & quadrant ABC .

If we join A to C , then it will be common for triangles ADC & ABC . And they will be congruent by SSS congruence condition.

Therefore the area of both quadrants will also be equal . Now we can find area of quadrant as ;

[tex]\large\boxed{\red{\bf Area_{quadrant}=\dfrac{\pi r^2}{4}}}[/tex]

Here radius will be equal to 5cm .

⇒ Area = πr² / 4 .

⇒ Area = π (5cm)² / 4 .

⇒ Area = 22/7 × 25 × 4 cm².

⇒ Area = 19.64 cm² .

So , total area of both quadrants = 39.28 cm² .

Also , area of square will be :

[tex]\large\boxed{\bf{\red{Area_{square}=(side)^2}}}[/tex]

⇒ Area = 5cm × 5cm .

⇒ Area = 25 cm².

Now , subtract area of one quadrant from the area of square = 25cm² - 19.64 cm² = 5.36 cm².

Similarly area of other white region = 5.36cm² .

And the areas sum will be = 5.36cm² × 2 = 10.72cm² .

Now , from the figure it's clear that ,

⇒ Area of unshaded region + Area of shaded region = 25cm².

⇒ 10.72cm² + ar( Shaded region ) = 25cm².

⇒ ar ( Shaded region ) = 25cm² - 10.72cm².

⇒ ar ( Shaded region ) = 14.28 cm².

Hence the area of shaded region is 14.28 cm².

[tex]\large\boxed{\red{\bf Answer = 14.28cm^2}}[/tex]

A process is normally distributed and in control, with known mean and variance, and the usual three-sigma limits are used on the control chart, so that the probability of a single point plotting outside the control limits when the process is in control is 0.0027. Suppose that this chart is being used in phase I and the averages from a set of m samples or subgroups from this process are plotted on this chart. What is the probability that at least one of the averages will plot outside the control limits when m = 5? Repeat these calculations for the cases where m = 10, m = 20, m = 30, and m = 50.

Required:
Discuss the results that you have obtained.

Answers

Answer:

The solution to the issue is outlined in the following portion of the summary.

Step-by-step explanation:

The given value is:

p = 0.0027

When m = 5,