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Ket [755]
3 years ago
15

Sara's mother has bought 90 lollipops and 105 pencils for goody bags for Sara's birthday. What is the largest number of goody ba

gs that Sara's mother can make so that each goody bag has the same number of lollipops and the same number of pencils?
Mathematics
1 answer:
crimeas [40]3 years ago
5 0

Answer:

<h3>15</h3>

Step-by-step explanation:

<h2>you need to find the lcf(least common factor) of the two numbers which is 15</h2>
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Anestetic [448]
Do 51×5 and you will get the answer 255
3 0
3 years ago
The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e
I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

\hat p=\frac{X}{n}=\frac{224}{269}=0.8327

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

1-\hat p=1-0.8327

        =0.1673

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

\hat q=1-\hat p

  =1-\frac{X}{n}

  =1-\frac{546}{709}

  =0.2299\\\approx 0.229

The critical value of <em>z</em> for 95% confidence interval is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Compute the 95% confidence interval for the population proportion as follows:

CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}

     =0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

5 0
3 years ago
F(x) = x^2-3<br> What is the average rate of change on the interval [-5, 2] ?
Mumz [18]

Answer:

(-3).

Step-by-step explanation:

Over this interval, the change in the value of this function is:

f(-5) - f(2) = 22 - 1 = 21.

The corresponding change in the value of x:

(-5) - 2 = -7.

The average rate of change of function f over this interval is equal to the change in the function value divided by the corresponding change in x:

\begin{aligned}& (\text{average rate of change}) \\ =\; & \frac{(\text{change in function value})}{(\text{change in $x$})} \\ =\; & \frac{f(-5) - f(2)}{(-5) - 2} \\ =\; & \frac{22 - 1}{-7} \\ =\; & \frac{21}{-7} \\ =\; & -3\end{aligned}.

Thus, the average rate of change of f(x) = x^{2} - 3 over the interval [-5,\, 2] would be (-3).

7 0
2 years ago
What's 28/42 simplified
Kryger [21]
28/42 ÷2
14/21 ÷2
2/3

28/42 in simplest form is = 2/3
8 0
3 years ago
Please please please help me I need it within 10 minutes!
Alja [10]
Mr Green- Brown tie
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I'm working on 6

Hope this helps in the meantime :)
5 0
3 years ago
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