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bija089 [108]
3 years ago
9

Jasper plans to serve cheese and crackers as an appetizer at his next dinner party. He buys 0.8 pounds of cheddar cheese. If the

cheese costs $5.35 per pound, how much does he spend?
Mathematics
2 answers:
Mashutka [201]3 years ago
7 0

Answer:

It might be 42.80

Step-by-step explanation:

5.35x0.8

azamat3 years ago
6 0

Answer: $4.28

Step-by-step explanation: Hey man,  you would multiply 0.8 pounds with $5.35. 0.8 times 5.35 = 4.28. He spent $4.28.

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It says that between any two real numbers there’s always another real number Rachel numbers any number that can be written in fraction form is a Rachel number this includes integers termination of decimals and repeating decimal as well as fractions so any termination decimal is the ratio number
6 0
2 years ago
Helppp me plsssssssss<br><br>​
Oliga [24]

Answer:

The class 35 - 40 has maximum frequency. So, it is the modal class.

From the given data,

  • \sf \:\:\:\:\:\:\:\:\:\:x_{k}=35
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k}=50
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k-1}=34
  • \sf \:\:\:\:\:\:\:\:\:\:f_{k+1}=42
  • \sf \:\:\:\:\:\:\:\:\:\:h=5

{\bf \:\: {By\:using\:the\: formula}} \\ \\

\:\dag\:{\small{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}}} \\ \\

\sf \:\:\:\:\:\:\:\:\:= 35+ {\bigg(5 \times \dfrac{(50 - 34)}{ ( 2 \times 50 - 34 - 42)}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= 35 +{\bigg(5 \times \dfrac{16}{24}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:= {\bigg(35+\dfrac{10}{3}\bigg)} \\ \\

\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(35 + 3.33) =.38.33 \\ \\

\:\:\sf {Hence,}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\ \large{\underline{\mathcal{\gray{ mode\:=\:38.33}}}} \\ \\

{\large{\frak{\pmb{\underline{Additional\: information }}}}}

MODE

  • Most precisely, mode is that value of the variable at which the concentration of the data is maximum.

MODAL CLASS

  • In a frequency distribution the class having maximum frequency is called the modal class.

{\bf{\underline{Formula\:for\: calculating\:mode:}}} \\

{\underline{\boxed{\sf {Mode,\:M_{o} =\sf\red{x_k + {\bigg(h \times \: \dfrac{ ( f_k - f_{k-1})}{ (2f_k - f_{k - 1} - f_{k +1})}\bigg)}}}}}} \\ \\

Where,

\sf \small\pink{ \bigstar} \: x_{k}= lower\:limit\:of\:the\:modal\:class\:interval.

\small \blue{ \bigstar}\sf \: f_{k}=frequency\:of\:the\:modal\:class

\sf \small\orange{ \bigstar}\: f_{k-1}=frequency\:of\:the\:class\: preceding\:the\;modal\:class

\sf \small\green{ \bigstar}\: f_{k+1}=frequency\:of\:the\:class\: succeeding\:the\;modal\:class

\small \purple{ \bigstar}\sf \: h= width \:of\:the\:class\:interval

7 0
3 years ago
Which image shows the correct position of -2 1/2on the number line?
ohaa [14]
The answer to your question would be 'A'
5 0
3 years ago
Marine biologists have been studying the effects of acidification of the oceans on weights of male baluga whales in the Arctic O
PSYCHO15rus [73]

Answer:

c. 61.25 kg

Step-by-step explanation:

The margin of error in estimating the true mean weight of male baluga whales in the Artic Ocean.

a. 15.31 kg

b. 51.40 kg

c. 61.25 kg

d. 80.49 kg

Margin of Error Formula= z × Standard deviation/√n

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Standard deviation = 125kg

n = 16 samples

Margin of error= 1.96 × 125/√16

= 1.96 × 125/4

= 245/4

= 61.25kg

The margin of error in estimating the true mean weight of male baluga whales in the Artic Ocean is 61.25kg

6 0
3 years ago
Can someone help me with this work please
tangare [24]

Answer:

-See below

Step-by.step explanation:

In the first column, on the left of the vertical line, you place the first digit of the number , then on the second row on you place the second digit.

So on the second row you have the entries:

1 | 2 2 4 representing  12, 12 and 14.

On the first  row the 2 0ne digit numbers 3 and 8 are represented by

0 | 3 8.

Similarly the last 2 rows are:

2 |  0 1 3 6

3 |  4

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