we decide to accept We will make one condition. combine with if clause​

Answer:

The BCC (Blind Carbon Copy) method is the most common approach to send emails to multiple recipients at the same time. Emailing to multiple recipients using the BCC feature hides other recipients from the recipient, making it look like he is the sole recipient of the email

Answer:

3 and 32 as x and y respectively

Step-by-step explanation:

here i go plz mark me brainliest

Answer:

The value of missing frequency x = [tex]3[/tex] and missing frequency y = [tex]32[/tex].

Step-by-step explanation:

Given,

Marks:    [tex]0-10[/tex]    [tex]10-20[/tex]    [tex]20-30[/tex]    [tex]30-40[/tex]    [tex]40-50[/tex]    [tex]50-60[/tex]

N ( [tex]f_{i}[/tex] ):      [tex]10[/tex]             x             [tex]25[/tex]            [tex]20[/tex]            y              [tex]10[/tex]

N = The number of students

The median of the given data = [tex]36[/tex]

The sum of all the given frequencies, [tex]\sum f_{i}[/tex] = [tex]100[/tex]

To find:

• The value of the missing frequencies x and y

Now, if the median is [tex]36[/tex], then the median class is [tex]30-40[/tex].

Now, we have to find out the cumulative frequency:

Marks     Number of students ( [tex]f_{i}[/tex] )     Cumulative frequency

[tex]0-10[/tex]                  [tex]10[/tex]                                           [tex]10[/tex]

[tex]10-20[/tex]                 x                                         [tex]10+x[/tex]

[tex]20-30[/tex]                [tex]25[/tex]                                        [tex]35+x[/tex]

[tex]30-40[/tex]                [tex]20[/tex]                                        [tex]55+x[/tex]

[tex]40-50[/tex]                y                                       [tex]55+x+y[/tex]

[tex]50-60[/tex]                [tex]10[/tex]                                      [tex]65+x+y[/tex]

Total                 [tex]100[/tex]

Now, the total number of students [tex]\sum f_{i}[/tex] = [tex]100[/tex]

Therefore,

• [tex]10+x+25+20+y+10[/tex] = [tex]100[/tex]

• [tex]65+x+y = 100[/tex]
• [tex]x+y =35[/tex]             -------equation (1)

Now,

• Median = [tex]l+\frac{[\frac{N}{2}-c.f] }{f} \times h[/tex]

Here,

• l = The median class's lower limit = [tex]30[/tex]
• N = The total frequency of the given data = [tex]100[/tex]
• c.f = The cumulative frequency of the preceding class = [tex]35+x[/tex]
• f = The frequency of the median class = [tex]20[/tex]
• h = The class size/height = [tex]10[/tex]

Thus,

• Median = [tex]l+\frac{[\frac{N}{2}-c.f] }{f} \times h[/tex]
• [tex]36[/tex] = [tex]30+\frac{[\frac{100}{2}-(35+x)] }{20} \times 10[/tex]
• [tex]36[/tex] = [tex]30+\frac{[50-35-x)] }{2}[/tex]
• [tex]36[/tex] = [tex]30+\frac{[15-x)] }{2}[/tex]
• [tex]36[/tex] = [tex]\frac{60+15-x}{2}[/tex]
• [tex]75-x = 72[/tex]
• x = [tex]3[/tex]

Now, after putting the value of x in equation (1), we get:

• [tex]3+y =35[/tex]
• y = [tex]32[/tex]

Hence, the value of missing frequency x = [tex]3[/tex] and missing frequency y = [tex]32[/tex].

Answer:

This statement is False.

Credit sales of goods are entered in sales journal and cash sales transactions are to be recorded in cash book in the receipt side.

Answer:

the state memt is false

Explanation:

Notice that only credit sales of inventory and merchandise items are recorded in the sales journal. Cash sales of inventory are recorded in the cash receipts journal. Both cash and credit sales of non-inventory or merchandise are recorded in the general journal