According to your fitness guide, 10,00 steps equals how many miles approximately? Question 5 options:

Aliya deposited half as much money in a savings account earning 1.9% simple interest as she invested in a money market account t

yanalaym [24]

Let A be the amount of money that Aliya deposited in the savings account. Since A is half as much as money as she invested in a money market account, then the amount that she invested in the market account is 2A.

Express the interest that Aliya earned in terms of A. Set it equal to the amount of $297.60 and then solve for A.

Since the savings account gives 1.9% simple interest, the total amount of interest that she will earn from the savings account is 1.9% of A, which is equal to:

$\frac{1.9}{100}\times A$

Since the money market account gives 3.7% simple interest, the total amount of interest that she will earn from the money market account, is 3.7% of 2A, which is equal to:

$\frac{3.7}{100}\times2A$

Add both interests in terms of A and simplify the expression:

$\begin{gathered} \frac{1.9}{100}\times A+\frac{3.7}{100}\times2A \\ =\frac{1.9}{100}\times A+\frac{7.4}{100}\times A \\ =(\frac{1.9}{100}+\frac{7.4}{100})\times A \\ =\frac{1.9+7.4}{100}\times A \\ =\frac{9.3}{100}\times A \end{gathered}$

The expression (9.3/100)*A represents the total interest after one year. Then:

$\begin{gathered} \frac{9.3}{100}\times A=297.60 \\ \Rightarrow A=\frac{100}{9.3}\times297.60 \\ \Rightarrow A=\frac{100\times297.60}{9.3} \\ \Rightarrow A=\frac{29760}{9.3} \\ \Rightarrow A=3200 \end{gathered}$

Use the value of A to find the amount that was invested in the money market account:

$2A=2\times3200=6400$

Therefore, Aliya deposited 3200 in a savings account and 6400 in a money market account.

3 0

1 year ago

In space movies, spacecrafts explode and oxygen is needed for a fire to start. Why is it wrong? What would really happen?

Triss [41]

Answer:

Hey

The reason these "space movies" are wrong is because objects in "space" (not in a "space" craft) can't catch on fire because there is no air in "space".

6 0

3 years ago

A Carnot engine whose high-temperature reservoir is at 464 K has an efficiency of 25.0%. By how much should the temperature of t

nexus9112 [7]

Answer

given,

high temperature reservoir (T_c)= 464 K

efficiency of reservoir (ε)= 25 %

temperature to decrease = ?

increase in efficiency = 42 %

now, using equation

$\epsilon = 1 - \dfrac{T_C}{T_H}$

$0.25 = 1 - \dfrac{T_C}{464}$

$T_C= (1 - 0.25) \times 464$

$T_C= 0.75 \times 464$

T_C = 348 K

now,

if the efficiency is equal to 42$ = 0.42

$T_C= (1 - 0.42) \times 464$

$T_C= 0.58 \times 464$

$T_C= 269.12\ K$

8 0

3 years ago

A boat moves through the water of a river at 10m/s relative to the water, regardless of the boat ‘s direction . If the water in

katen-ka-za [31]

Answer:

The appropriate solution is "61.37 s".

Explanation:

The given values are:

Boat moves,

= 10 m/s

Water flowing,

= 1.50 m/s

Displacement,

d = 300 m

Now,

The boat is travelling,

= $10+1.50$

= $11.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v = \frac{d}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{11.5}$

$=26.08 \ s$

Throughout the opposite direction, when the boat seems to be travelling then,

= $10-1.50$

= $8.5 \ m/s$

Travelling such distance for 300 m will be:

⇒ $v=\frac{v}{t} \ sot \ t$

$=\frac{d}{v}$

On putting the values, we get

$=\frac{300}{8.5}$

$=35.29 \ s$

hence,

The time taken by the boat will be:

= $26.08+35.29$

= $61.37 \ s$

8 0

2 years ago

When you jump straight up as high as you can, what is the order of magnitude of the maximum recoil speed that you give to the Ea

Klio2033 [76]

Answer:

5.66 × 10⁻²³ m/s

Explanation:

If i assume i can jump as high as h = 2 m, my initial velocity is gotten from v² = u² + 2gh. Since my final velocity v = 0, u = √2gh = √(2 × 9.8 × 2) = √39.2 m/s = 6.26 m/s.

Since initial momentum = final momentum,

mv₁ + MV₁ = mv₂ + MV₂ where m, M, v₁, V₁, v₂ and V₂ are my mass, mass of earth, my initial velocity, earth's initial velocity, my final velocity and earth's final velocity respectively.

My mass m = 54 kg, M = 5.972 × 10²⁴ kg, v₁ = 6.26 m/s, V₁ = 0, v₂ = 0 and V₂ = ?

So mv₁ + M × 0 = m × 0 + MV₂

mv₁ = MV₂

V₂ = mv₁/M = 54kg × 6.26 m/s/5.972 × 10²⁴ kg = 338.093/5.972 × 10²⁴ = 56.61 × 10⁻²⁴ m/s = 5.661 × 10⁻²³ m/s ≅ 5.66 × 10⁻²³ m/s

5 0

3 years ago