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puteri [66]
3 years ago
11

Alex buys 24 apples and when he opens it 6 apples are bad whats the perecentage of the bad apples

Mathematics
1 answer:
nataly862011 [7]3 years ago
6 0
6/24=1/4=25%

answer is 25% are bad
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Write the next three terms of the arithmetic sequence.
Inga [223]

Answer:

36, 44, 52

Step-by-step explanation:

Just add 8 to 28 = 36

Add 8 to 36 = 44

Add 8 to 44 = 52

8 0
2 years ago
Read 2 more answers
(5+22) x (35-27) + 6^2
aksik [14]
What does ^ stand for?
7 0
3 years ago
2. Solve: 33 + 2(5+8)-(6-2)
mars1129 [50]

Answer:

55

Step-by-step explanation:

Add the numbers 5 + 8 and you get 13. Multiply 13 x 2 which is 26. Add 33 + 26 and you get 59. Then, subtract it by 4 and you get 55.

6 0
3 years ago
Joni has a circular garden with a diameter of 14 1/2 feet if she uses 2 teaspoons of fertilizer for every 25 square feet of gard
alina1380 [7]

Answer:

13.20 teaspoon of fertiliser needed for entire garden.

Step-by-step explanation:

Diameter of the garden = 14.50 ft

Radius of the garden = 7.25 ft

Area

[tex]A=\pi r^2[/text]

[tex]A=\pi \times 7.25 \times 7.25[/text]

[tex]\pi=3.14[/text]

[tex]A=\pi \times 7.25 \times 7.25[/text]

[tex]A=3.14 \times 7.25 \times 7.25[/text]

Hence

[tex]A=165.046[/text]

Area is 165.046 Sq Ft

For 25 sq ft of fields the fertiliser needed = 2 tea spoon

For 1 sq ft of fields the fertiliser needed =

[tex]\frac{2}{25}[/text] tea spoon

For 165.046 sq ft of fields the fertiliser needed =

[tex]\frac{2 \times 165.046 }{25}[/text] tea spoon

For 165.046 sq ft of fields the fertiliser needed = 13.20 tea spoon

6 0
2 years ago
Does a statistics course improve a student's mathematics skills,as measured by a national test? Suppose a random sample of 13 st
viva [34]

Answer:

There is not enough evidence to support the claim that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

P-value=0.042.

Step-by-step explanation:

The question is incomplete:

The data of the scores for each student is:

Before    After

430        465

485        475

520        535

360        410

440        425

500        505

425        450

470        480

515        520

430        430

450        460

495        500

540        530

We will generate a sample for the difference of scores (before - after) and test that sample.

The sample of the difference is [35 -10 15 50 -15 5 25 10 5 0 10 5 -10]

This sample, of size n=13, has a mean of 9.615 and a standard deviation of 18.423.

The claim is that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

Then, the null and alternative hypothesis are:

H_0: \mu=0\\\\H_a:\mu> 0

The significance level is 0.01.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{18.423}{\sqrt{13}}=5.11

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{9.615-0}{5.11}=\dfrac{9.615}{5.11}=1.882

The degrees of freedom for this sample size are:

df=n-1=13-1=12

This test is a right-tailed test, with 12 degrees of freedom and t=1.882, so the P-value for this test is calculated as (using a t-table):

P-value=P(t>1.882)=0.042

As the P-value (0.042) is bigger than the significance level (0.01), the effect is  not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the scores after the stats course are significantly higher than the scores before (the difference in the scores is higher than 0).

4 0
2 years ago
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