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3 years ago

4. The cost of pizzas at two different shops is shown in the graph and table

Mathematics

1 answer:

ruslelena [56]3 years ago

7 0

Answer:

I am not too sure but i think 6+5=11

Step-by-step explanation:

where 6 comes first then 11 comes second and that equals 11

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What is the value of s and the length of side BC if ABCD is a rhombus?

Natali5045456 [20]

Step 1

Find the value of s

we know that

In a Rhombus all sides are congruent

$AB=BC=CD=DA$

$AB=9s+29\\CD=10s-16$

equate AB and CD

$9s+29=10s-16\\$

Combine like term

$10s-9s=29+16\\$

$s=45\ units$

The answer Part a) is

the value of s is $45\ units$

Step 2

Find the value of side AB

$AB=9s+29$

substitute the value of s

$AB=9*45+29=434\ units$

Remember that the sides are congruent

$AB=BC=CD=DA$

therefore

the answer Part b) is

The length of the side BC is $434\ units$

4 0

3 years ago

Read 2 more answers

Convert 3 4/7 into an improper fraction.

Marizza181 [45]

25/7
You multiply 3 and 7 which gives you 21. Add 4 to 21, giving you 25. The improper fraction would be 25/7

7 0

3 years ago

All the fourth-graders in a certain elementary school took a standardized test. A total of 85% of the students were found to be

Aneli [31]

Answer:

There is a 2% probability that the student is proficient in neither reading nor mathematics.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a student is proficient in reading

B is the probability that a student is proficient in mathematics.

C is the probability that a student is proficient in neither reading nor mathematics.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a student is proficient in reading but not mathematics and $A \cap B$ is the probability that a student is proficient in both reading and mathematics.

By the same logic, we have that:

$B = b + (A \cap B)$

Either a student in proficient in at least one of reading or mathematics, or a student is proficient in neither of those. The sum of the probabilities of these events is decimal 1. So

$(A \cup B) + C = 1$