Answer:
She needs at least 11 pieces of accessories.
Step-by-step explanation:
She needs $200 and she has $45
200 - 45 = 155
Now she needs $155 worth of accessories and each piece is worth $15
155/15 = 10.33
then you round up to find the answer of 11 pieces of accessories.
Answer:
length = 11 ft, width = 6 ft
Step-by-step explanation:
Let w be the width, l be the length and p the perimeter. You can write down the following equations:
1. w = l - 5
2. p = 2*(w + l)
Now you have two equations with two unknown.
Plugging eq.1 in eq.2:
p = 2*(l-5 + l)
p = 2*(2*l - 5)
p = 4*l - 10
Remember p = 34
34 = 4*l - 10
4*l = 44
l = 11
Using eq.1 to solve for w:
w = 11 - 5
w = 6
We can see the ratio of a side big figure and small figure is 60:15 or 60/15 = 1/4.
In decimal form we can write 1/4 as 0.25.
Therefore, scale factor is 0.25.
So, second option is applicable. Please check second option :
<h3>The scale factor is 0.25.</h3>
Now, let us find the side A by multiplying by scale factor 0.25 of the big figure.
The corresponding side of A is 35.
Therefore, A is 0.25 × 35 = 8.75 cm.
Therefore, other applicable option is 3rd option.
<h3>The length of side A is 8.75 cm.</h3>
She will pay 20$ for 10 pounds of hamburger
Answer:
The sample consisting of 64 data values would give a greater precision.
Step-by-step explanation:
The width of a (1 - <em>α</em>)% confidence interval for population mean <em>μ</em> is:

So, from the formula of the width of the interval it is clear that the width is inversely proportion to the sample size (<em>n</em>).
That is, as the sample size increases the interval width would decrease and as the sample size decreases the interval width would increase.
Here it is provided that two different samples will be taken from the same population of test scores and a 95% confidence interval will be constructed for each sample to estimate the population mean.
The two sample sizes are:
<em>n</em>₁ = 25
<em>n</em>₂ = 64
The 95% confidence interval constructed using the sample of 64 values will have a smaller width than the the one constructed using the sample of 25 values.
- Width for <em>n</em> = 25:
- Width for <em>n</em> = 64:
![\text{Width}=2\cdot z_{\alpha/2}\cdot \frac{\sigma}{\sqrt{64}}=\frac{1}{8}\ [2\cdot z_{\alpha/2}\cdot \sigma]](https://tex.z-dn.net/?f=%5Ctext%7BWidth%7D%3D2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7B64%7D%7D%3D%5Cfrac%7B1%7D%7B8%7D%5C%20%5B2%5Ccdot%20z_%7B%5Calpha%2F2%7D%5Ccdot%20%5Csigma%5D)
Thus, the sample consisting of 64 data values would give a greater precision.