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3 years ago

Grant has $75 to spend on gifts for his daughter. He buys dolls for $15 each and toys for $8 each.

Mathematics

2 answers:

Fantom [35]3 years ago

8 0

Answer:

A. 15d+8t≤75

Step-by-step explanation:

Since D represents the amount of money he can spend on dolls and T represents the amount of money he can spend on toys, and he can only spend less than or equal to $75 on the gifts

deff fn [24]3 years ago

4 0

Answer:

Step-by-step explanation:

Your going to want the money to equal less then $75 dollars not more because if it more than the amount you have you cant buy it.

Hope this helps :3

(feedback helps me! dont be afraid to tell me if im wrong)<3

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Help please! Thank you!!

umka2103 [35]

Answer:

i dont know what it is

Step-by-step explanation:

maybe a

3 0

3 years ago

Q5: The probability that event A occurs is 5/7 and the probability that event B occurs is 2/3 . If A and B are independent event

bekas [8.4K]

Answer:

$P(A \cap B)$

And we can use the following formula:

$P(A \cap B)= P(A)* P(B)$

And replacing the info we got:

$P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476$

Step-by-step explanation:

We define two events for this case A and B. And we know the probability for each individual event given by the problem:

$p(A) = \frac{5}{7}$

$p(B) = \frac{2}{3}$

And we want to find the probability that A and B both occurs if A and B are independent events, who menas the following conditions:

$P(A|B) = P(A)$

$P(B|A) = P(B)$

And for this special case we want to find this probability:

$P(A \cap B)$

And we can use the following formula:

$P(A \cap B)= P(A)* P(B)$

And replacing the info we got:

$P(A \cap B) = \frac{5}{7} \frac{2}{3}= \frac{10}{21}=0.476$

7 0

3 years ago

Pls pls i need help with this

Scrat [10]

Answer:

5+8=13

5/13

Lisa's fraction is 5/13

6 0

3 years ago

Please help quickly i need this done

mr Goodwill [35]

222, 223, 224, 225, 233, 233, 234, 235, 242 243, 244, 245, 252, 253, 254, 255

322, 323, 324, 325, 332, 333, 334, 335, 342, 343, 344, 345, 352, 353, 354, 355

422, 423, 424, 425, 432, 433, 434, 435, 442, 443, 444, 445, 452, 453, 454, 455

522, 523, 524, 525, 532, 533, 534, 535, 542, 543, 544, 545, 552, 553, 554, 555

So, as you see, here are all the possiblities. The ones in bold are the odd numbers that can be formed without any repeating digits. There are 13 of those, so your answer should be 13.

Hope this helped! Have a great rest of your day! :)

8 0

3 years ago

Matthew currently has enough money to buy 45 toy cars. If the cost of each car was 10 cents less, Matthew could buy 5 more cars.

Law Incorporation [45]

The answer is $45

Let's make the system of equations.
x - the cost of the car
y - the amount Matthew has

Matthew currently has enough money to buy 45 toy cars:
y =45x
If the cost of each car was 10 cents less, Matthew could buy 5 more cars:
y = (x - 10)*(45 + 5)

45x = (x - 0.10) * 50
45x = 50x - 5
5 = 50x - 45x
5 = 5x
x = 5 / 5
x = 1

y = 45x = 45 * 1 = $45

5 0

3 years ago