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3 years ago

7

Martin made two type of cookies.He used 2/3 cup of sugar for one recipe and 1/4 cup for the other one. How much sugar did he use

in all?

2 answers:

Ymorist [56]3 years ago

4 0

.916666666....

or 11/12.

kiruha [24]3 years ago

3 0

You must add 2/3 and 1/4. You need a common denominator. The least common denominator of 3 and 4 is 12.

2/3 + 1/4 =

= 2/3 * 4/4 + 1/4 * 3/3

= 8/12 + 3/12

= 11/12

Answer: He used 11/12 of a cup of sugar.

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Evaluate t (-4). T(z)=z-6

jekas [21]

Simply plug -4 in for z in the equation. The result should give you T(z)=(-4)-6=-10
Notice -4 takes z spot

8 0

3 years ago

A collectible car has been going up in price by 10% each year. if the car is already worth $29,000, how much will the car be

sammy [17]

The worth of the collectible car in five years if its value is going up by 10% every year would be $46,704.79

<h3>What is the worth of the car?</h3>

If the value of the collectible car is increasing with time, it is said that the collectible car is appreciating in value.

The formula for calculating future value with appreciation:

FV = P (1 + r)^n

FV = Future value
P = Present value
R = interest rate
N = number of years

$29,000(1.1^5) = $46,704.79

To learn more about future value, please check: brainly.com/question/18760477

3 0

2 years ago

A well known pharmaceutical manufacturer is manufacturing a newly developed vaccine, and is concerned about variability in the i

Lana71 [14]

Answer:

a

The null hypothesis is $H_o : \ \sigma^2 = 1.9$

The alternative hypothesis is $H_a : \ \sigma^2 \ne 1.9$

b

$X^2 = 13.74$

c

The decision rule is

Reject the null hypothesis

d

The conclusion is

There is no sufficient evidence to conclude that the variance of the immune response is equal to 1.9.

Step-by-step explanation:

From the question we are are told that

The variance is $\sigma ^2 = 1.9$

The sample size is n = 30

The sample variance is $s^2 = 0.9$

The level of significance is $\alpha = 0.05$

The null hypothesis is $H_o : \ \sigma^2 = 1.9$

The alternative hypothesis is $H_a : \ \sigma^2 \ne 1.9$

Generally the test statistics is mathematically represented as

$X^2 = \frac{ (n- 1 ) * s^2 }{ \sigma^2 }$

=> $X^2 = \frac{ (30 - 1 ) * 0.9 }{1.9 }$

=> $X^2 = 13.74$

Generally from the degree of freedom is mathematically represented as

$df = n- 1$

=> $df = 30 - 1$

=> $df = 29$

Generally from the chi distribution table the critical value of $\frac{\alpha}{2} \ and \ 1 - \frac{ \alpha }{2}$ at a degree of freedom of $df = 29$ is

$X^2 _{ \frac{\alpha }{2} , df } = X^2 _{ \frac{0.05 }{2} , 29 } = 45.7$

=> $X^2 _{ 1 - \frac{\alpha }{2} , df } = X^2 _{1 - \frac{0.05 }{2} , 29 } = 16.0 5$

Gnerally from the values obtained we see that

$X^2 < X^2 _{ 1 - \frac{\alpha }{2} , df }$ hence

The decision rule is

Reject the null hypothesis

The conclusion is

There is no sufficient evidence to conclude that the variance of the immune response is equal to 1.9.

6 0

3 years ago

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Paraphin [41]

1

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3 years ago

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Nutka1998 [239]

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2 years ago