All categories

2 months ago

A grapefruit is 8% heavier than an orange, and an apple is 10% lighter than the orange.

2 answers:

8 0

The weight of a grapefruit equals the weight of an orange plus 8% of the weight of an orange
The weight of an apple equals the weight of an orange minus 10% of the weight of an orange

a.What percentage heavier is the grapefruit compared to the apple?
To determine this, we should establish the relationship between the grapefruit and the apple. Knowing the grapefruit’s relation to an orange and the orange’s relation to an apple indicates the connection between a grapefruit and an apple.

The weight of a grapefruit = the weight of an orange + 8% of the weight of an orangeThe weight of an orange = the weight of an apple + 10% of the weight of an apple

-> The weight of a grapefruit = the weight of an apple + 10% of the weight of an apple + 8%(weight of an apple + 10% of the weight of an apple) = weight of an apple + 18% of the weight of an apple + 2% of the weight of an apple = weight of an apple + 20% of the weight of an apple

b. By what percentage is the apple lighter than the grapefruit?
The weight of a grapefruit = the weight of an apple + 20% of the weight of an apple

-> The apple is 20% lighter than the grapefruit.

zzz [12.3K]2 months ago

3 0

Response:

A. 20%

B. 16 2/3%

Detailed explanation:

24

Step-by-step explanation:

Based on the logarithmic expressions given $log_ax = 3, log_ay = 7, log_az = -2$ , we need to identify the value of $log_a(\frac{x^3y}{z^4} )$

$from\ log_ax = 3, x = a^3\\\\from\ log_ay = 7,y = a^7\\\\from\ log_az = -2, z = a^{-2}$ By substituting x = a³, y = a⁷, and z = a⁻² into the logarithmic function $log_a(\frac{x^3y}{z^4} )$ , we will derive;

$= log_a(\frac{x^3y}{z^4} )\\\\= log_a(\dfrac{(a^3)^3*a^7}{(a^{-2})^4} )\\\\= log_a(\dfrac{a^9*a^7}{a^{-8}} )\\\\= log_a\dfrac{a^{16}}{a^{-8}} \\\\= log_aa^{16+8}\\\\= log_aa^{24}\\\\= 24log_aa\\\\= 24* 1\\\\= 24$

Therefore, the result of the logarithmic expression is 24

6 0

1 month ago

An engineer wishes to determine the width of a particular electronic component. If she knows that the standard deviation of all

Leona [12618]

Answer:

$n=(\frac{0.539(2.2)}{0.3})^2 =15.62 \approx 16$

In this scenario, the solution for n is determined to be 16 when rounded up to the closest integer

Step-by-step explanation:

The following information has been provided for this case:

$\sigma = 2.2$ symbolizes the population standard deviation

$Confidence =0.8$

$ME = 0.3$ signifies the preferred margin of error

The formula for the margin of error of the actual mean is given as:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

In this case, the margin of error (ME) is 0.03, and to find n, we rearrange equation (a) to get:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

With a confidence level of 80%, and the significance level being $\alpha=0.2$ and $\alpha/2 =0.1$ , the critical value is $z_{\alpha/2}=0.539$ . Plugging in these values into formula (b) results in:

$n=(\frac{0.539(2.2)}{0.3})^2 =15.62 \approx 16$

Thus, the solution for this situation results in n=16, rounded to the nearest whole number

5 0

1 month ago

Compute P7,2. (Enter an exact number.)

Zina [12379]

42. The permutation formula is P(n, r) = n! / (n - r)!. Given n = 7 and r = 2, we have: 7! / (7 - 2)! = 7! / 5!. This simplifies to 7 * 6 (since 5! cancels out), resulting in 42.

8 0

1 month ago

Read 2 more answers