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3 years ago

Jade made 6 cups of blueberry jam and didvided the jam equally among 4 containers.how much jam went in each container

Mathematics

1 answer:

never [62]3 years ago

8 0

Answer: 3/2 of jam went in each container

Step-by-step explanation:

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If line XZ is 2ft, line YZ=18in. what is the length of line XY

Lunna [17]

Answer:

XY = 6 in

Step-by-step explanation:

XZ = 2ft = 24 in.

XY + YZ = 24 in.

XY + 18 in. = 24 in.

XY = 24 - 18 = 6 in

3 0

3 years ago

PLEASE I NED SOMEONE TO HELP ME WITH THIS QUESTION

tigry1 [53]

Answer:

a) 12 points

b) A: 3 questions; B: 2 questions

Step-by-step explanation:

The least number of points each section may have is the least common multiple of 4 and 6. That is a number that will have all of the factors of 4 and all of the factors of 6: 2×2×3 = 12.

Each part is worth 12 points.

Part A has 12/4 = 3 questions.

Part B has 12/6 = 2 questions.

8 0

2 years ago

Determine whether the sampling is dependent or independent. Indicate whether the response variable is qualitative or quantitativ

Papessa [141]

Answer:

Dependent.

Quantitative.

Step-by-step explanation:

<u>Determine whether the sampling is dependent or independent.</u>

This is a matched-pair data test, as the same students are measured two times in order to compare the results.

This type of samples are dependent, as they are correlated between them.

<u>Indicate whether the response variable is qualitative or quantitative.</u>

The response variable is the test score. This is a quantitative variable, as this can be defined by a number. Both sample scores are averaged and compared.

6 0

3 years ago

En la tienda de mascotas "Animalo-T", se desea elevar un elefante de 2,900 kg utilizando una elevadora hidráulica de plato grand

Korvikt [17]

Answer:

Se requiere una fuerza de 854.473 newtons sobre el émbolo pequeño.

Step-by-step explanation:

Por el Principio de Pascal se conoce que el esfuerzo experimentado por el elefante es igual a la presión ejercida por el plato pequeño. Es decir:

$\frac{F_{1}}{A_{1}} = \frac{F_{2}}{A_{2}}$ (1)

Donde:

$F_{1}$ - Fuerza experimentada por el elefante, medida en newtons.

$F_{2}$ - Fuerza aplicada sobre el plato pequeño, medida en newtons.

$A_{1}$ - Área del plato grande, medida en metros cuadrados.

$A_{2}$ - Área del plato pequeño, medida en metros cuadrados.

La fuerza aplicada sobre el plato pequeño es:

$F_{2} = \left(\frac{A_{2}}{A_{1}} \right)\cdot F_{1}$

La fuerza experimentada por el elefante es su propio peso. Por otra parte, el área del plato es directamente proporcional al cuadrado de su diámetro. Es decir:

$F_{2} = \left(\frac{D_{2}}{D_{1}} \right)^{2}\cdot m\cdot g$ (2)

Donde:

$D_{1}$ - Diámetro del plato grande, medido en centímetros.

$D_{2}$ - Diámetro del plato pequeño, medido en centímetros.

$m$ - Masa del elefante, medida en kilogramos.

$g$ - Aceleración gravitacional, medida en metros por segundo cuadrado.

Si sabemos que $D_{1} = 0.75\,m$ , $D_{2} = 0.13\,m$ , $m = 2900\,kg$ y $g = 9.807\,\frac{m}{s^{2}}$ , entonces la fuerza a aplicar al émbolo pequeño es: