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3 years ago

$15.30 for 3 small pizzas

Mathematics

2 answers:

stellarik [79]3 years ago

8 0

Answer: $5.10

Explanation: $15.30 divided by 3 = $5.10

BabaBlast [244]3 years ago

5 0

Answer:

$5.1

Step-by-step explanation:

If it costs $15.30 for 3 pizzas, then we can divide 15.30 by 3 to get $5.1. This concludes that the cost for one pizza is $5.1.

Hope this helps!!! PLZ MARK BRAINLIEST!!!

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Item 5 There are 20 balls in a bag. Ten of the balls are red, 4 are green, and the rest are yellow. Which ratio represents the n

OlgaM077 [116]

Answer:

3:5 the colon is needed to represent the ratio

Step-by-step explanation:

r = 10

g = 4

y = 30 - 14 = 6

y = 6

so ratio of total balls is 10:4:6

and ratio of y:r is

6:10

simply by dividing by 2

3:5

4 0

3 years ago

Which of the values shown are potential roots of f(x) = 3x3 – 13x2 – 3x + 45? Select all that apply.

galina1969 [7]

Answer:

All potential roots are 3,3 and $-\frac{5}{3}$ .

Step-by-step explanation:

Potential roots of the polynomial is all possible roots of f(x).

$f(x)=3x^3-13x^2-3x+45$

Using rational root theorem test. We will find all the possible or potential roots of the polynomial.

p=All the positive/negative factors of 45

q=All the positive/negative factors of 3

$p=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45$

$q=\pm 1,\pm 3$

All possible roots

$\frac{p}{q}=\pm 1,\pm 3,\pm 5\pm \pm 9,\pm 15\pm 45,\pm \frac{1}{3},\pm \frac{5}{3}$

Now we check each rational root and see which are possible roots for given function.

$f(1)= 3\times 1^3-13\times 1^2-3\times 1+45\Rightarrow 32\neq 0$

$f(-1)= 3\times (-1)^3-13\times (-1)^2-3\times (-1)+45\Rightarrow \neq 32$

$f(-3)= 3\times (-3)^3-13\times (-3)^2-3\times (-3)+45\Rightarrow \neq -144$

$f(3)= 3\times (3)^3-13\times (3)^2-3\times (3)+45\Rightarrow =0\\\\ \therefore x=3\text{ Potential roots of function}$

Similarly, we will check for all value of p/q and we get

$f(-5/3)=0$

Thus, All potential roots are 3,3 and $-\frac{5}{3}$ .

5 0

2 years ago

Read 2 more answers

Joseph saved his money to buy your table computer he purchased a computer with those cuz she had he gave the sale clark 4 hundre

andre [41]

Hello! We know that we have $400. So 23 ten dollar bills is equal to $230. How? We multiply 23 x 10 and get $230. Then, we have 46 dimes. A dime is worth 10 cents so we multiply 46 x 10 which equals $4.60. And of course we have the 19 cents. Now we add. 400 + 230 + 4.60 + 19 = $653.6.

Hope this Helps!

4 0

3 years ago

Using the traditional formula, a 95% CI for p1 − p2 is to be constructed based on equal sample sizes from the two populations. F

inn [45]

Answer:

Step-by-step explanation:

Given:

Width = 0.4

Let's take the z value of 95% which is = 1.96

Let's assume population proportion p1 and p2 = 0.5

For margin of error E, the relation between the width and margin of error is 2E.

i.e 2E = 0.4

E = 0.2

To find the sample size, n, let's use the formula :

$n= \frac{z^2*(p_1(1-p_1) + p_2(1-p_2))}{E^2}$

$= \frac{1.96^2(0.5(1-0.5)+0.5(1-0.5))}{0.2^2}$

= 48.02

≈ 49

8 0

3 years ago

Mystery Boxes: Breakout Rooms

ollegr [7]

Answer:

$\begin{array}{ccccccccccccccc}{1} & {3} & {4} & {12} & {15} & {18}& {18 } & {26} & {29} & {30} & {32} & {57} & {58} & {61} \\ \end{array}$

Step-by-step explanation:

Given

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {[ \ ]} \\ \end{array}$

Required

Fill in the box

From the question, the range is:

$Range = 60$

Range is calculated as:

$Range = Highest - Least$

From the box, we have:

$Least = 1$

So:

$60 = Highest - 1$

$Highest = 60 +1$

$Highest = 61$

The box, becomes:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

From the question:

$IQR = 20$ --- interquartile range

This is calculated as:

$IQR = Q_3 - Q_1$

$Q_3$ is the median of the upper half while $Q_1$ is the median of the lower half.

So, we need to split the given boxes into two equal halves (7 each)

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {[ \ ] } & {15} & {18}& {[ \ ] } \\ \end{array}$

Upper half

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

The quartile is calculated by calculating the median for each of the above halves is calculated as:

$Median = \frac{N + 1}{2}th$

Where N = 7

So, we have:

$Median = \frac{7 + 1}{2}th = \frac{8}{2}th = 4th$

So,

$Q_3$ = 4th item of the upper halves

$Q_1$ = 4th item of the lower halves

From the upper halves

$\begin{array}{ccccccc}{[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$

We have:

$Q_3 = 32$

$Q_1$ can not be determined from the lower halves because the 4th item is missing.

So, we make use of:

$IQR = Q_3 - Q_1$

Where $Q_3 = 32$ and $IQR = 20$

So:

$20 = 32 - Q_1$

$Q_1 = 32 - 20$

$Q_1 = 12$

So, the lower half becomes:

Lower half:

$\begin{array}{ccccccc}{1} & {[ \ ]} & {4} & {12 } & {15} & {18}& {[ \ ] } \\ \end{array}$

From this, the updated values of the box is:

$\begin{array}{ccccccccccccccc}{1} & {[ \ ]} & {4} & {12} & {15} & {18}& {[ \ ] } & {[ \ ]} & {29} & {30} & {32} & {[ \ ]} & {58} & {61} \\ \end{array}$