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3 years ago

60 questions i need 70% how many can i get erong

Mathematics

1 answer:

GREYUIT [131]3 years ago

5 0

42/60 = 70% exactly.
The answer: At least 42.

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What is the solution of the inequality shown

Colt1911 [192]

Hi!

To solve this, we have to isolate the variable on one side of the equation.

Given:

$b+10>-1$

Subtract both sides by $10$

$b+10(-10)$

Your answer is:

$b$

4 0

2 years ago

Mrs white wants to crochet beach hits and baby afghans for a church fund raising bazaar. she needs 7 hours to make a hat and 4 h

Georgia [21]

She can make 5 hats and 6 afghans in 60 hours and will make $159 and have a good stock of each item or she could make 8 hats and 1 afghan in 60 hours and make $204 for the most profit.

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3 years ago

The weight of an object rounds to 41.3 kilograms select the three posible weights that correctly round to 41.3 kilograms

nekit [7.7K]

Answer:

See Explanation

Step-by-step explanation:

Given

$Weight = 41.3kg$

Required

Determine the weights that approximates to 41.3kg

The options are not given but the question is still solvable

In approximation:

0 - 4 approximates to 0

This implies that:

41.30 to 41.34 approximates to 41.3

Also:

5 - 9 approximates to 1

This implies that:

41.25 to 41.29 approximates to 41.3

Bring the range together, we have:

$Range = 41.25\ to\ 41.34$

Select any three numbers within this range, and you have your answer

Examples are: 41.25, 41.28, 41.34\

4 0

4 years ago

9. On a game show, there are 16 questions: 8 easy, 5 medium-hard, and 3 hard. If contestants are given questions

just olya [345]

Answer:

$\frac{7}{30}\approx0.23$

Step-by-step explanation:

Given:

Number of questions (N) = 16

Number of easy questions (E) = 8

Number of medium-hard questions (M) = 5

Number of hard questions (H) = 3

Now, the probability of getting the first question as easy question is given as:

$P(E1)=\frac{\textrm{Number of easy questions}}{\textrm{Total questions}}\\\\P(E1)=\frac{E}{N}=\frac{8}{16}=0.5$

Now, probability of getting the second question as easy question is given as:

$P(E2)=\frac{\textrm{Number of easy questions left}}{\textrm{Total questions left}}\\\\P(E2)=\frac{E-1}{N-1}=\frac{7}{15}$

Now, probability that the first two contestants will get easy questions is given by the product of $P(E1)\ and\ P(E2)$ . So,

$P(2\ easy\ questions)=P(E1)\times P(E2)\\\\P(2\ easy\ questions)=\frac{1}{2}\times \frac{7}{15}\\\\P(2\ easy\ questions)=\frac{7}{30}\approx 0.23$

Therefore, the probability that the first two contestants will get easy questions is $\frac{7}{30}\approx0.23$

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3 years ago

Find the indicated term of the arithmetic sequence.

dmitriy555 [2]

Answer:

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers